Chapter 10 of 14

Updated 6 August 2026

NCERT Solutions

Conic Sections — NCERT Solutions

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CBSE · Class 11 · Mathematics

NCERT Solutions for Conic Sections, CBSE Class 11 Mathematics: 70 textbook questions solved step by step. Part of the CBSE Class 11 Mathematics syllabus.

193 questions56 flashcards14 formulas & key relations5 concepts

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Exercise 10.1

1Find the equation of the circle with centre

(0,2)

and radius

2

.Show solution

Given: Centre $(h,k) = (0,2)$ , radius $r = 2$ .

Formula: $(x-h)^2 + (y-k)^2 = r^2$

Solution:
$(x-0)^2 + (y-2)^2 = 2^2$
$x^2 + (y-2)^2 = 4$

Expanding:
$x^2 + y^2 - 4y + 4 = 4$
$\boxed{x^2 + y^2 - 4y = 0}$

2Find the equation of the circle with centre

(-2,3)

and radius

4

.Show solution

Given: Centre $(h,k) = (-2,3)$ , radius $r = 4$ .

Formula: $(x-h)^2 + (y-k)^2 = r^2$

Solution:
$(x-(-2))^2 + (y-3)^2 = 4^2$
$(x+2)^2 + (y-3)^2 = 16$

Expanding:
$x^2 + 4x + 4 + y^2 - 6y + 9 = 16$
$\boxed{x^2 + y^2 + 4x - 6y - 3 = 0}$

3Find the equation of the circle with centre

\left(\dfrac{1}{2}, \dfrac{1}{4}\right)

and radius

\dfrac{1}{12}

.Show solution

Given: Centre $(h,k) = \left(\dfrac{1}{2}, \dfrac{1}{4}\right)$ , radius $r = \dfrac{1}{12}$ .

Formula: $(x-h)^2 + (y-k)^2 = r^2$

Solution:
$\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \left(\frac{1}{12}\right)^2$

$\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144}$

Expanding:
$x^2 - x + \frac{1}{4} + y^2 - \frac{y}{2} + \frac{1}{16} = \frac{1}{144}$

Multiplying throughout by $144$ :
$144x^2 - 144x + 36 + 144y^2 - 72y + 9 = 1$
$\boxed{144x^2 + 144y^2 - 144x - 72y + 44 = 0}$

4Find the equation of the circle with centre

(1,1)

and radius

\sqrt{2}

.Show solution

Given: Centre $(h,k) = (1,1)$ , radius $r = \sqrt{2}$ .

Formula: $(x-h)^2 + (y-k)^2 = r^2$

Solution:
$(x-1)^2 + (y-1)^2 = (\sqrt{2})^2$
$(x-1)^2 + (y-1)^2 = 2$

Expanding:
$x^2 - 2x + 1 + y^2 - 2y + 1 = 2$
$\boxed{x^2 + y^2 - 2x - 2y = 0}$

5Find the equation of the circle with centre

(-a,-b)

and radius

\sqrt{a^2 - b^2}

.Show solution

Given: Centre $(h,k) = (-a,-b)$ , radius $r = \sqrt{a^2 - b^2}$ .

Formula: $(x-h)^2 + (y-k)^2 = r^2$

Solution:
$(x-(-a))^2 + (y-(-b))^2 = \left(\sqrt{a^2-b^2}\right)^2$
$(x+a)^2 + (y+b)^2 = a^2 - b^2$

Expanding:
$x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 - b^2$
$\boxed{x^2 + y^2 + 2ax + 2by + 2b^2 = 0}$

6Find the centre and radius of the circle

(x+5)^2 + (y-3)^2 = 36

.Show solution

Given: $(x+5)^2 + (y-3)^2 = 36$

Concept: Comparing with standard form $(x-h)^2 + (y-k)^2 = r^2$ :

$x - h = x + 5 \Rightarrow h = -5$
$y - k = y - 3 \Rightarrow k = 3$
$r^2 = 36 \Rightarrow r = 6$

Centre $= (-5, 3)$ , Radius $= 6$ .

7Find the centre and radius of the circle

x^2 + y^2 - 4x - 8y - 45 = 0

.Show solution

Given: $x^2 + y^2 - 4x - 8y - 45 = 0$

Method: Complete the square.

$(x^2 - 4x) + (y^2 - 8y) = 45$
$(x^2 - 4x + 4) + (y^2 - 8y + 16) = 45 + 4 + 16$
$(x-2)^2 + (y-4)^2 = 65$

Comparing with $(x-h)^2 + (y-k)^2 = r^2$ :

Centre $= (2, 4)$ , Radius $= \sqrt{65}$ .

8Find the centre and radius of the circle

x^2 + y^2 - 8x + 10y - 12 = 0

.Show solution

Given: $x^2 + y^2 - 8x + 10y - 12 = 0$

Method: Complete the square.

$(x^2 - 8x) + (y^2 + 10y) = 12$
$(x^2 - 8x + 16) + (y^2 + 10y + 25) = 12 + 16 + 25$
$(x-4)^2 + (y+5)^2 = 53$

Comparing with $(x-h)^2 + (y-k)^2 = r^2$ :

Centre $= (4, -5)$ , Radius $= \sqrt{53}$ .

9Find the centre and radius of the circle

2x^2 + 2y^2 - x = 0

.Show solution

Given: $2x^2 + 2y^2 - x = 0$

Divide throughout by $2$ :
$x^2 + y^2 - \frac{x}{2} = 0$

Method: Complete the square.
$\left(x^2 - \frac{x}{2}\right) + y^2 = 0$
$\left(x^2 - \frac{x}{2} + \frac{1}{16}\right) + y^2 = \frac{1}{16}$
$\left(x - \frac{1}{4}\right)^2 + (y-0)^2 = \frac{1}{16}$

Comparing with $(x-h)^2 + (y-k)^2 = r^2$ :

Centre $= \left(\dfrac{1}{4}, 0\right)$ , Radius $= \dfrac{1}{4}$ .

10Find the equation of the circle passing through the points

(4,1)

and

(6,5)

and whose centre is on the line

4x + y = 16

.Show solution

Given: Circle passes through $(4,1)$ and $(6,5)$ ; centre lies on $4x + y = 16$ .

Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$ .

Step 1: Since $(4,1)$ lies on the circle:
$(4-h)^2 + (1-k)^2 = r^2 \quad \cdots(1)$

Step 2: Since $(6,5)$ lies on the circle:
$(6-h)^2 + (5-k)^2 = r^2 \quad \cdots(2)$

Step 3: Centre lies on $4x + y = 16$ :
$4h + k = 16 \quad \cdots(3)$

Step 4: From (1) = (2):
$(4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2$
$16 - 8h + h^2 + 1 - 2k + k^2 = 36 - 12h + h^2 + 25 - 10k + k^2$
$17 - 8h - 2k = 61 - 12h - 10k$
$4h + 8k = 44$
$h + 2k = 11 \quad \cdots(4)$

Step 5: Solving (3) and (4):
From (3): $k = 16 - 4h$
Substituting in (4): $h + 2(16-4h) = 11$
$h + 32 - 8h = 11 \Rightarrow -7h = -21 \Rightarrow h = 3$
$k = 16 - 12 = 4$

Step 6: Find $r^2$ using point $(4,1)$ :
$r^2 = (4-3)^2 + (1-4)^2 = 1 + 9 = 10$

Equation of circle:
$(x-3)^2 + (y-4)^2 = 10$
$\boxed{x^2 + y^2 - 6x - 8y + 15 = 0}$

11Find the equation of the circle passing through the points

(2,3)

and

(-1,1)

and whose centre is on the line

x - 3y - 11 = 0

.Show solution

Given: Circle passes through $(2,3)$ and $(-1,1)$ ; centre lies on $x - 3y - 11 = 0$ .

Let the equation be $(x-h)^2 + (y-k)^2 = r^2$ .

Step 1: Since $(2,3)$ lies on the circle:
$(2-h)^2 + (3-k)^2 = r^2 \quad \cdots(1)$

Step 2: Since $(-1,1)$ lies on the circle:
$(-1-h)^2 + (1-k)^2 = r^2 \quad \cdots(2)$

Step 3: Centre on $x - 3y - 11 = 0$ :
$h - 3k - 11 = 0 \quad \cdots(3)$

Step 4: From (1) = (2):
$(2-h)^2 + (3-k)^2 = (1+h)^2 + (1-k)^2$
$4 - 4h + h^2 + 9 - 6k + k^2 = 1 + 2h + h^2 + 1 - 2k + k^2$
$13 - 4h - 6k = 2 + 2h - 2k$
$11 = 6h + 4k$
$6h + 4k = 11 \quad \cdots(4)$

Step 5: Solving (3) and (4):
From (3): $h = 3k + 11$
Substituting in (4): $6(3k+11) + 4k = 11$
$18k + 66 + 4k = 11 \Rightarrow 22k = -55 \Rightarrow k = -\frac{5}{2}$
$h = 3\left(-\frac{5}{2}\right) + 11 = -\frac{15}{2} + 11 = \frac{7}{2}$

Step 6: Find $r^2$ using point $(2,3)$ :
$r^2 = \left(2 - \frac{7}{2}\right)^2 + \left(3 + \frac{5}{2}\right)^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{11}{2}\right)^2 = \frac{9}{4} + \frac{121}{4} = \frac{130}{4} = \frac{65}{2}$

Equation of circle:
$\left(x - \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{65}{2}$

Expanding and multiplying by 1:
$x^2 - 7x + \frac{49}{4} + y^2 + 5y + \frac{25}{4} = \frac{65}{2}$
$x^2 + y^2 - 7x + 5y + \frac{74}{4} - \frac{130}{4} = 0$
$\boxed{x^2 + y^2 - 7x + 5y - 14 = 0}$

12Find the equation of the circle with radius 5 whose centre lies on

x

-axis and passes through the point

(2,3)

.Show solution

Given: Radius $r = 5$ ; centre on $x$ -axis; circle passes through $(2,3)$ .

Since centre lies on $x$ -axis, let centre $= (h, 0)$ .

Step 1: Since the circle passes through $(2,3)$ :
$(2-h)^2 + (3-0)^2 = 5^2$
$(2-h)^2 + 9 = 25$
$(2-h)^2 = 16$
$2 - h = \pm 4$
$h = 2 - 4 = -2 \quad \text{or} \quad h = 2 + 4 = 6$

Step 2: Two possible circles:

Centre $(-2, 0)$ : $(x+2)^2 + y^2 = 25$ , i.e., $x^2 + y^2 + 4x - 21 = 0$
Centre $(6, 0)$ : $(x-6)^2 + y^2 = 25$ , i.e., $x^2 + y^2 - 12x + 11 = 0$

$\boxed{x^2 + y^2 + 4x - 21 = 0 \quad \text{or} \quad x^2 + y^2 - 12x + 11 = 0}$

13Find the equation of the circle passing through

(0,0)

and making intercepts

a

and

b

on the coordinate axes.Show solution

Given: Circle passes through origin $(0,0)$ , makes intercept $a$ on $x$ -axis and $b$ on $y$ -axis.

Step 1: Since the circle makes intercept $a$ on $x$ -axis, it passes through $(a, 0)$ .
Since it makes intercept $b$ on $y$ -axis, it passes through $(0, b)$ .

Let the equation be $(x-h)^2 + (y-k)^2 = r^2$ .

Step 2: Passes through $(0,0)$ :
$h^2 + k^2 = r^2 \quad \cdots(1)$

Step 3: Passes through $(a,0)$ :
$(a-h)^2 + k^2 = r^2 \quad \cdots(2)$

Step 4: Passes through $(0,b)$ :
$h^2 + (b-k)^2 = r^2 \quad \cdots(3)$

Step 5: From (1) and (2):
$(a-h)^2 + k^2 = h^2 + k^2$
$a^2 - 2ah = 0 \Rightarrow h = \frac{a}{2}$

Step 6: From (1) and (3):
$h^2 + (b-k)^2 = h^2 + k^2$
$b^2 - 2bk = 0 \Rightarrow k = \frac{b}{2}$

Step 7: $r^2 = h^2 + k^2 = \dfrac{a^2}{4} + \dfrac{b^2}{4}$

Equation:
$\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4}$

Expanding:
$x^2 - ax + \frac{a^2}{4} + y^2 - by + \frac{b^2}{4} = \frac{a^2+b^2}{4}$
$\boxed{x^2 + y^2 - ax - by = 0}$

14Find the equation of a circle with centre

(2,2)

and passes through the point

(4,5)

.Show solution

Given: Centre $(h,k) = (2,2)$ ; circle passes through $(4,5)$ .

Step 1: Find radius:
$r = \sqrt{(4-2)^2 + (5-2)^2} = \sqrt{4 + 9} = \sqrt{13}$

Step 2: Equation of circle:
$(x-2)^2 + (y-2)^2 = 13$

Expanding:
$x^2 - 4x + 4 + y^2 - 4y + 4 = 13$
$\boxed{x^2 + y^2 - 4x - 4y - 5 = 0}$

15Does the point

(-2.5, 3.5)

lie inside, outside or on the circle

x^2 + y^2 = 25

?Show solution

Given: Point $(-2.5, 3.5)$ ; circle $x^2 + y^2 = 25$ (centre $O=(0,0)$ , radius $r = 5$ ).

Step 1: Find the distance from the centre to the point:
$d = \sqrt{(-2.5)^2 + (3.5)^2} = \sqrt{6.25 + 12.25} = \sqrt{18.5}$

Step 2: Compare with radius:
$\sqrt{18.5} \approx 4.3 < 5 = r$

Since $d < r$ , the point lies inside the circle.

Alternatively: Substitute in $x^2 + y^2$ :
$(-2.5)^2 + (3.5)^2 = 6.25 + 12.25 = 18.5 < 25$

Since $18.5 < 25$ , the point $(-2.5, 3.5)$ lies inside the circle.

Exercise 10.2

1Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of

y^2 = 12x

.Show solution

Given: $y^2 = 12x$

Comparing with $y^2 = 4ax$ :
$4a = 12 \Rightarrow a = 3$

Focus: $(a, 0) = (3, 0)$
Axis: $x$ -axis (i.e., $y = 0$ )
Directrix: $x = -a$ , i.e., $x = -3$
Length of latus rectum: $4a = 12$

2Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of

x^2 = 6y

.Show solution

Given: $x^2 = 6y$

Comparing with $x^2 = 4ay$ :
$4a = 6 \Rightarrow a = \frac{3}{2}$

Focus: $\left(0, \dfrac{3}{2}\right)$
Axis: $y$ -axis (i.e., $x = 0$ )
Directrix: $y = -\dfrac{3}{2}$
Length of latus rectum: $4a = 6$

3Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of

y^2 = -8x

.Show solution

Given: $y^2 = -8x$

Comparing with $y^2 = -4ax$ :
$4a = 8 \Rightarrow a = 2$

Focus: $(-a, 0) = (-2, 0)$
Axis: $x$ -axis (i.e., $y = 0$ )
Directrix: $x = a = 2$ , i.e., $x = 2$
Length of latus rectum: $4a = 8$

4Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of

x^2 = -16y

.Show solution

Given: $x^2 = -16y$

Comparing with $x^2 = -4ay$ :
$4a = 16 \Rightarrow a = 4$

Focus: $(0, -a) = (0, -4)$
Axis: $y$ -axis (i.e., $x = 0$ )
Directrix: $y = a = 4$ , i.e., $y = 4$
Length of latus rectum: $4a = 16$

5Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of

y^2 = 10x

.Show solution

Given: $y^2 = 10x$

Comparing with $y^2 = 4ax$ :
$4a = 10 \Rightarrow a = \frac{5}{2}$

Focus: $\left(\dfrac{5}{2}, 0\right)$
Axis: $x$ -axis (i.e., $y = 0$ )
Directrix: $x = -\dfrac{5}{2}$
Length of latus rectum: $4a = 10$

6Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of

x^2 = -9y

.Show solution

Given: $x^2 = -9y$

Comparing with $x^2 = -4ay$ :
$4a = 9 \Rightarrow a = \frac{9}{4}$

Focus: $\left(0, -\dfrac{9}{4}\right)$
Axis: $y$ -axis (i.e., $x = 0$ )
Directrix: $y = \dfrac{9}{4}$
Length of latus rectum: $4a = 9$

7Find the equation of the parabola with Focus

(6,0)

; directrix

x = -6

.Show solution

Given: Focus $(6,0)$ , directrix $x = -6$ .

Since focus is on the $x$ -axis and directrix is $x = -6$ , the parabola is of the form $y^2 = 4ax$ with $a = 6$ .

$\boxed{y^2 = 24x}$

8Find the equation of the parabola with Focus

(0,-3)

; directrix

y = 3

.Show solution

Given: Focus $(0,-3)$ , directrix $y = 3$ .

Since focus is on the $y$ -axis (negative side) and directrix is $y = 3$ , the parabola opens downward: $x^2 = -4ay$ with $a = 3$ .

$\boxed{x^2 = -12y}$

9Find the equation of the parabola with Vertex

(0,0)

; focus

(3,0)

.Show solution

Given: Vertex $(0,0)$ , focus $(3,0)$ .

Focus lies on positive $x$ -axis, so parabola is of the form $y^2 = 4ax$ with $a = 3$ .

$\boxed{y^2 = 12x}$

10Find the equation of the parabola with Vertex

(0,0)

; focus

(-2,0)

.Show solution

Given: Vertex $(0,0)$ , focus $(-2,0)$ .

Focus lies on negative $x$ -axis, so parabola is of the form $y^2 = -4ax$ with $a = 2$ .

$\boxed{y^2 = -8x}$

11Find the equation of the parabola with Vertex

(0,0)

passing through

(2,3)

and axis is along

x

-axis.Show solution

Given: Vertex $(0,0)$ , passes through $(2,3)$ , axis along $x$ -axis.

Since axis is along $x$ -axis and vertex is at origin, the equation is either $y^2 = 4ax$ or $y^2 = -4ax$ .

Since the point $(2,3)$ has $x > 0$ , the parabola opens to the right: $y^2 = 4ax$ .

Substituting $(2,3)$ :
$9 = 4a(2) \Rightarrow a = \frac{9}{8}$

$y^2 = 4 \cdot \frac{9}{8} \cdot x$
$\boxed{2y^2 = 9x}$

12Find the equation of the parabola with Vertex

(0,0)

, passing through

(5,2)

and symmetric with respect to

y

-axis.Show solution

Given: Vertex $(0,0)$ , passes through $(5,2)$ , symmetric about $y$ -axis.

Since symmetric about $y$ -axis and vertex at origin, equation is $x^2 = 4ay$ or $x^2 = -4ay$ .

Since $(5,2)$ has $y > 0$ , parabola opens upward: $x^2 = 4ay$ .

Substituting $(5,2)$ :
$25 = 4a(2) \Rightarrow a = \frac{25}{8}$

$x^2 = 4 \cdot \frac{25}{8} \cdot y = \frac{25}{2}y$
$\boxed{2x^2 = 25y}$

Exercise 10.3

1Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1

.Show solution

Given: $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$

Here $a^2 = 36$ , $b^2 = 16$ , so $a = 6$ , $b = 4$ . Since $a^2 > b^2$ and the larger denominator is under $x^2$ , the major axis is along the $x$ -axis.

$c = \sqrt{a^2 - b^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}$

Foci: $(\pm 2\sqrt{5},\, 0)$
Vertices: $(\pm 6,\, 0)$
Length of major axis: $2a = 12$
Length of minor axis: $2b = 8$
Eccentricity: $e = \dfrac{c}{a} = \dfrac{2\sqrt{5}}{6} = \dfrac{\sqrt{5}}{3}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 16}{6} = \dfrac{16}{3}$

2Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

\dfrac{x^2}{4} + \dfrac{y^2}{25} = 1

.Show solution

Given: $\dfrac{x^2}{4} + \dfrac{y^2}{25} = 1$

Here $a^2 = 25$ , $b^2 = 4$ (since $25 > 4$ , major axis is along $y$ -axis), $a = 5$ , $b = 2$ .

$c = \sqrt{25 - 4} = \sqrt{21}$

Foci: $(0,\, \pm\sqrt{21})$
Vertices: $(0,\, \pm 5)$
Length of major axis: $2a = 10$
Length of minor axis: $2b = 4$
Eccentricity: $e = \dfrac{\sqrt{21}}{5}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{5} = \dfrac{8}{5}$

3Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1

.Show solution

Given: $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$

Here $a^2 = 16$ , $b^2 = 9$ , $a = 4$ , $b = 3$ . Major axis along $x$ -axis.

$c = \sqrt{16 - 9} = \sqrt{7}$

Foci: $(\pm\sqrt{7},\, 0)$
Vertices: $(\pm 4,\, 0)$
Length of major axis: $2a = 8$
Length of minor axis: $2b = 6$
Eccentricity: $e = \dfrac{\sqrt{7}}{4}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 9}{4} = \dfrac{9}{2}$

4Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

\dfrac{x^2}{25} + \dfrac{y^2}{100} = 1

.Show solution

Given: $\dfrac{x^2}{25} + \dfrac{y^2}{100} = 1$

Here $a^2 = 100$ , $b^2 = 25$ (major axis along $y$ -axis), $a = 10$ , $b = 5$ .

$c = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3}$

Foci: $(0,\, \pm 5\sqrt{3})$
Vertices: $(0,\, \pm 10)$
Length of major axis: $2a = 20$
Length of minor axis: $2b = 10$
Eccentricity: $e = \dfrac{5\sqrt{3}}{10} = \dfrac{\sqrt{3}}{2}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 25}{10} = 5$

5Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

\dfrac{x^2}{49} + \dfrac{y^2}{36} = 1

.Show solution

Given: $\dfrac{x^2}{49} + \dfrac{y^2}{36} = 1$

Here $a^2 = 49$ , $b^2 = 36$ , $a = 7$ , $b = 6$ . Major axis along $x$ -axis.

$c = \sqrt{49 - 36} = \sqrt{13}$

Foci: $(\pm\sqrt{13},\, 0)$
Vertices: $(\pm 7,\, 0)$
Length of major axis: $2a = 14$
Length of minor axis: $2b = 12$
Eccentricity: $e = \dfrac{\sqrt{13}}{7}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 36}{7} = \dfrac{72}{7}$

6Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

\dfrac{x^2}{100} + \dfrac{y^2}{400} = 1

.Show solution

Given: $\dfrac{x^2}{100} + \dfrac{y^2}{400} = 1$

Here $a^2 = 400$ , $b^2 = 100$ (major axis along $y$ -axis), $a = 20$ , $b = 10$ .

$c = \sqrt{400 - 100} = \sqrt{300} = 10\sqrt{3}$

Foci: $(0,\, \pm 10\sqrt{3})$
Vertices: $(0,\, \pm 20)$
Length of major axis: $2a = 40$
Length of minor axis: $2b = 20$
Eccentricity: $e = \dfrac{10\sqrt{3}}{20} = \dfrac{\sqrt{3}}{2}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 100}{20} = 10$

7Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

36x^2 + 4y^2 = 144

.Show solution

Given: $36x^2 + 4y^2 = 144$

Dividing by $144$ : $\dfrac{x^2}{4} + \dfrac{y^2}{36} = 1$

Here $a^2 = 36$ , $b^2 = 4$ (major axis along $y$ -axis), $a = 6$ , $b = 2$ .

$c = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$

Foci: $(0,\, \pm 4\sqrt{2})$
Vertices: $(0,\, \pm 6)$
Length of major axis: $2a = 12$
Length of minor axis: $2b = 4$
Eccentricity: $e = \dfrac{4\sqrt{2}}{6} = \dfrac{2\sqrt{2}}{3}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{6} = \dfrac{4}{3}$

8Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

16x^2 + y^2 = 16

.Show solution

Given: $16x^2 + y^2 = 16$

Dividing by $16$ : $\dfrac{x^2}{1} + \dfrac{y^2}{16} = 1$

Here $a^2 = 16$ , $b^2 = 1$ (major axis along $y$ -axis), $a = 4$ , $b = 1$ .

$c = \sqrt{16 - 1} = \sqrt{15}$

Foci: $(0,\, \pm\sqrt{15})$
Vertices: $(0,\, \pm 4)$
Length of major axis: $2a = 8$
Length of minor axis: $2b = 2$
Eccentricity: $e = \dfrac{\sqrt{15}}{4}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 1}{4} = \dfrac{1}{2}$

9Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

4x^2 + 9y^2 = 36

Given: $4x^2 + 9y^2 = 36$

Dividing by $36$ : $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$

Here $a^2 = 9$ , $b^2 = 4$ (major axis along $x$ -axis), $a = 3$ , $b = 2$ .

$c = \sqrt{9 - 4} = \sqrt{5}$

Foci: $(\pm\sqrt{5},\, 0)$
Vertices: $(\pm 3,\, 0)$
Length of major axis: $2a = 6$
Length of minor axis: $2b = 4$
Eccentricity: $e = \dfrac{\sqrt{5}}{3}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{3} = \dfrac{8}{3}$

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10Find the equation for the ellipse with Vertices

(\pm 5, 0)

, foci

(\pm 4, 0)

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11Find the equation for the ellipse with Vertices

(0, \pm 13)

, foci

(0, \pm 5)

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12Find the equation for the ellipse with Vertices

(\pm 6, 0)

, foci

(\pm 4, 0)

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13Find the equation for the ellipse with ends of major axis

(\pm 3, 0)

, ends of minor axis

(0, \pm 2)

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14Find the equation for the ellipse with ends of major axis

(0, \pm\sqrt{5})

, ends of minor axis

(\pm 1, 0)

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15Find the equation for the ellipse with length of major axis

26

, foci

(\pm 5, 0)

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16Find the equation for the ellipse with length of minor axis

16

, foci

(0, \pm 6)

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17Find the equation for the ellipse with foci

(\pm 3, 0)

a = 4

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18Find the equation for the ellipse with

b = 3

c = 4

, centre at the origin; foci on the

x

-axis.

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19Find the equation for the ellipse with centre at

(0,0)

, major axis on the

y

-axis and passes through the points

(3,2)

and

(1,6)

Given: Major axis on $y$ -axis; passes through $(3,2)$ and $(1,6)$ .

Let the equation be $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ (major axis along $y$ -axis).

Step 1: Substituting $(3,2)$ :
$\frac{9}{b^2} + \frac{4}{a^2} = 1 \quad \cdots(1)$

Step 2: Substituting $(1,6)$ :
$\frac{1}{b^2} + \frac{36}{a^2} = 1 \quad \cdots(2)$

Step 3: Let $u = \dfrac{1}{b^2}$ , $v = \dfrac{1}{a^2}$ :
$9u + 4v = 1 \quad \cdots(1)$
$u + 36v = 1 \quad \cdots(2)$

From (2): $u = 1 - 36v$
Substituting in (1): $9(1-36v) + 4v = 1$
$9 - 324v + 4v = 1 \Rightarrow -320v = -8 \Rightarrow v = \frac{1}{40}$
$u = 1 - 36 \times \frac{1}{40} = 1 - \frac{9}{10} = \frac{1}{10}$

So $b^2 = 10$ , $a^2 = 40$ .

$\boxed{\frac{x^2}{10} + \frac{y^2}{40} = 1}$

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20Find the equation for the ellipse with major axis on the

x

-axis and passes through the points

(4,3)

and

(6,2)

Given: Major axis on $x$ -axis; passes through $(4,3)$ and $(6,2)$ .

Let the equation be $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ .

Step 1: Substituting $(4,3)$ :
$\frac{16}{a^2} + \frac{9}{b^2} = 1 \quad \cdots(1)$

Step 2: Substituting $(6,2)$ :
$\frac{36}{a^2} + \frac{4}{b^2} = 1 \quad \cdots(2)$

Step 3: Let $p = \dfrac{1}{a^2}$ , $q = \dfrac{1}{b^2}$ :
$16p + 9q = 1 \quad \cdots(1)$
$36p + 4q = 1 \quad \cdots(2)$

From (1): $q = \dfrac{1-16p}{9}$
Substituting in (2): $36p + 4 \cdot \dfrac{1-16p}{9} = 1$
$324p + 4 - 64p = 9 \Rightarrow 260p = 5 \Rightarrow p = \frac{1}{52}$
$q = \frac{1 - 16/52}{9} = \frac{36/52}{9} = \frac{36}{468} = \frac{1}{13}$

So $a^2 = 52$ , $b^2 = 13$ .

$\boxed{\frac{x^2}{52} + \frac{y^2}{13} = 1}$

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Exercise 10.4

1Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1

Given: $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$

Here $a^2 = 16$ , $b^2 = 9$ , so $a = 4$ , $b = 3$ . Transverse axis along $x$ -axis.

$c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Foci: $(\pm 5,\, 0)$
Vertices: $(\pm 4,\, 0)$
Eccentricity: $e = \dfrac{c}{a} = \dfrac{5}{4}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 9}{4} = \dfrac{9}{2}$

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2Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

\dfrac{y^2}{9} - \dfrac{x^2}{27} = 1

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3Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

9y^2 - 4x^2 = 36

Given: $9y^2 - 4x^2 = 36$

Dividing by $36$ : $\dfrac{y^2}{4} - \dfrac{x^2}{9} = 1$

Here $a^2 = 4$ , $b^2 = 9$ , $a = 2$ , $b = 3$ . Transverse axis along $y$ -axis.

$c = \sqrt{4 + 9} = \sqrt{13}$

Foci: $(0,\, \pm\sqrt{13})$
Vertices: $(0,\, \pm 2)$
Eccentricity: $e = \dfrac{\sqrt{13}}{2}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 9}{2} = 9$

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4Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

16x^2 - 9y^2 = 576

Given: $16x^2 - 9y^2 = 576$

Dividing by $576$ : $\dfrac{x^2}{36} - \dfrac{y^2}{64} = 1$

Here $a^2 = 36$ , $b^2 = 64$ , $a = 6$ , $b = 8$ . Transverse axis along $x$ -axis.

$c = \sqrt{36 + 64} = \sqrt{100} = 10$

Foci: $(\pm 10,\, 0)$
Vertices: $(\pm 6,\, 0)$
Eccentricity: $e = \dfrac{10}{6} = \dfrac{5}{3}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 64}{6} = \dfrac{64}{3}$

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5Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

5y^2 - 9x^2 = 36

Given: $5y^2 - 9x^2 = 36$

Dividing by $36$ : $\dfrac{y^2}{36/5} - \dfrac{x^2}{4} = 1$

Here $a^2 = \dfrac{36}{5}$ , $b^2 = 4$ , $a = \dfrac{6}{\sqrt{5}}$ , $b = 2$ . Transverse axis along $y$ -axis.

$c = \sqrt{\frac{36}{5} + 4} = \sqrt{\frac{36 + 20}{5}} = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}}$

Foci: $\left(0,\, \pm\dfrac{2\sqrt{14}}{\sqrt{5}}\right)$ , i.e., $\left(0,\, \pm\dfrac{2\sqrt{70}}{5}\right)$
Vertices: $\left(0,\, \pm\dfrac{6}{\sqrt{5}}\right)$ , i.e., $\left(0,\, \pm\dfrac{6\sqrt{5}}{5}\right)$
Eccentricity: $e = \dfrac{c}{a} = \dfrac{2\sqrt{14}/\sqrt{5}}{6/\sqrt{5}} = \dfrac{2\sqrt{14}}{6} = \dfrac{\sqrt{14}}{3}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{6/\sqrt{5}} = \dfrac{8\sqrt{5}}{6} = \dfrac{4\sqrt{5}}{3}$

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6Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

49y^2 - 16x^2 = 784

Given: $49y^2 - 16x^2 = 784$

Dividing by $784$ : $\dfrac{y^2}{16} - \dfrac{x^2}{49} = 1$

Here $a^2 = 16$ , $b^2 = 49$ , $a = 4$ , $b = 7$ . Transverse axis along $y$ -axis.

$c = \sqrt{16 + 49} = \sqrt{65}$

Foci: $(0,\, \pm\sqrt{65})$
Vertices: $(0,\, \pm 4)$
Eccentricity: $e = \dfrac{\sqrt{65}}{4}$
Length of latus rectum: $\dfrac{2b^2}{a} = \dfrac{2 \times 49}{4} = \dfrac{49}{2}$

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7Find the equation of the hyperbola with vertices

(\pm 2, 0)

, foci

(\pm 3, 0)

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8Find the equation of the hyperbola with vertices

(0, \pm 5)

, foci

(0, \pm 8)

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9Find the equation of the hyperbola with vertices

(0, \pm 3)

, foci

(0, \pm 5)

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10Find the equation of the hyperbola with foci

(\pm 5, 0)

, the transverse axis is of length

8

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11Find the equation of the hyperbola with foci

(0, \pm 13)

, the conjugate axis is of length

24

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12Find the equation of the hyperbola with foci

(\pm 3\sqrt{5}, 0)

, the latus rectum is of length

8

Given: Foci $(\pm 3\sqrt{5}, 0)$ , latus rectum $= 8$ .

$c = 3\sqrt{5}$ . Transverse axis along $x$ -axis.

Latus rectum $= \dfrac{2b^2}{a} = 8 \Rightarrow b^2 = 4a$ .

Also $c^2 = a^2 + b^2$ :
$45 = a^2 + 4a$
$a^2 + 4a - 45 = 0$
$(a+9)(a-5) = 0 \Rightarrow a = 5 \quad (\text{taking positive value})$
$b^2 = 4 \times 5 = 20$

$\boxed{\frac{x^2}{25} - \frac{y^2}{20} = 1}$

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13Find the equation of the hyperbola with foci

(\pm 4, 0)

, the latus rectum is of length

12

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14Find the equation of the hyperbola with vertices

(\pm 7, 0)

e = \dfrac{4}{3}

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15Find the equation of the hyperbola with foci

(0, \pm\sqrt{10})

, passing through

(2,3)

Given: Foci $(0, \pm\sqrt{10})$ , passes through $(2,3)$ .

Transverse axis along $y$ -axis: $c = \sqrt{10}$ .

Let equation be $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ .

$c^2 = a^2 + b^2 \Rightarrow a^2 + b^2 = 10 \Rightarrow b^2 = 10 - a^2 \quad \cdots(1)$

Substituting $(2,3)$ :
$\frac{9}{a^2} - \frac{4}{b^2} = 1 \quad \cdots(2)$

Substituting $b^2 = 10 - a^2$ in (2):
$\frac{9}{a^2} - \frac{4}{10 - a^2} = 1$
$9(10 - a^2) - 4a^2 = a^2(10 - a^2)$
$90 - 9a^2 - 4a^2 = 10a^2 - a^4$
$a^4 - 23a^2 + 90 = 0$

Let $t = a^2$ : $t^2 - 23t + 90 = 0$
$(t - 18)(t - 5) = 0 \Rightarrow t = 5 \text{ or } t = 18$

If $a^2 = 18$ : $b^2 = 10 - 18 = -8 < 0$ (rejected).
If $a^2 = 5$ : $b^2 = 10 - 5 = 5$ .

$\boxed{\frac{y^2}{5} - \frac{x^2}{5} = 1}$
i.e., $y^2 - x^2 = 5$

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Miscellaneous Exercise on Chapter 10

1If a parabolic reflector is

20

cm in diameter and

5

cm deep, find the focus.

Given: Diameter $= 20$ cm, depth $= 5$ cm.

Setup: Place the vertex of the parabola at the origin with axis along positive $x$ -axis. Equation: $y^2 = 4ax$ .

The reflector is $5$ cm deep, so the point on the rim has $x = 5$ .
Diameter $= 20$ cm, so the rim point is at $y = 10$ (half of diameter).

Step 1: Substituting $(5, 10)$ in $y^2 = 4ax$ :
$100 = 4a(5) = 20a \Rightarrow a = 5$

Focus is at $(a, 0) = (5, 0)$ , i.e., 5 cm from the vertex.

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2An arch is in the form of a parabola with its axis vertical. The arch is

10

m high and

5

m wide at the base. How wide is it

2

m from the vertex of the parabola?

Given: Arch height $= 10$ m, width at base $= 5$ m.

Setup: Place vertex at origin, axis along negative $y$ -axis (opens downward): $x^2 = -4ay$ .

The base is at $y = -10$ (10 m below vertex), and width $= 5$ m, so the base point is $\left(\dfrac{5}{2}, -10\right)$ .

Step 1: Substituting $\left(\dfrac{5}{2}, -10\right)$ :
$\frac{25}{4} = -4a(-10) = 40a \Rightarrow a = \frac{25}{160} = \frac{5}{32}$

Equation: $x^2 = -4 \cdot \dfrac{5}{32} \cdot y = -\dfrac{5}{8}y$

Step 2: At $2$ m from vertex, $y = -2$ :
$x^2 = -\frac{5}{8}(-2) = \frac{10}{8} = \frac{5}{4}$
$x = \frac{\sqrt{5}}{2}$

Width $= 2x = 2 \times \dfrac{\sqrt{5}}{2} = \sqrt{5}$ m $\approx 2.23$ m.

The arch is $\sqrt{5}$ m wide at $2$ m from the vertex.

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3The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and

100

m long is supported by vertical wires attached to the cable, the longest wire being

30

m and the shortest being

6

m. Find the length of a supporting wire attached to the roadway

18

m from the middle.

Given: Roadway length $= 100$ m; longest wire $= 30$ m; shortest wire $= 6$ m.

Setup: Place origin at the lowest point of the cable (where shortest wire is). The parabola opens upward: $x^2 = 4ay$ .

The shortest wire (at centre) $= 6$ m. The roadway extends $50$ m on each side.
At $x = 50$ , the wire length $= 30$ m, so the cable height above road $= 30 - 6 = 24$ m.
Thus point $(50, 24)$ lies on the parabola.

Step 1: Substituting $(50, 24)$ :
$2500 = 4a(24) \Rightarrow a = \frac{2500}{96} = \frac{625}{24}$

Step 2: At $x = 18$ :
$y = \frac{x^2}{4a} = \frac{324}{4 \times 625/24} = \frac{324 \times 24}{2500} = \frac{7776}{2500} = 3.1104$

Length of wire $= y + 6 = 3.1104 + 6 = 9.11$ m $\approx 9.11$ m.

The supporting wire at $18$ m from the middle is approximately $9.11$ m long.

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4An arch is in the form of a semi-ellipse. It is

8

m wide and

2

m high at the centre. Find the height of the arch at a point

1.5

m from one end.

Given: Width $= 8$ m, height at centre $= 2$ m.

Setup: Place centre of ellipse at origin, major axis along $x$ -axis.
$a = 4$ (semi-major), $b = 2$ (semi-minor).

Equation: $\dfrac{x^2}{16} + \dfrac{y^2}{4} = 1$

Step 1: A point $1.5$ m from one end means $x = 4 - 1.5 = 2.5$ from centre.

Step 2: Find $y$ at $x = 2.5$ :
$\frac{(2.5)^2}{16} + \frac{y^2}{4} = 1$
$\frac{6.25}{16} + \frac{y^2}{4} = 1$
$\frac{y^2}{4} = 1 - \frac{6.25}{16} = \frac{9.75}{16}$
$y^2 = \frac{39}{16} \Rightarrow y = \frac{\sqrt{39}}{4} \approx \frac{6.245}{4} \approx 1.56 \text{ m}$

The height of the arch at $1.5$ m from one end is $\dfrac{\sqrt{39}}{4} \approx 1.56$ m.

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5A rod of length

12

cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point

P

on the rod, which is

3

cm from the end in contact with the

x

-axis.

Given: Rod of length $12$ cm; ends on coordinate axes; $P$ is $3$ cm from the end on $x$ -axis.

Setup: Let the end on $x$ -axis be $A$ and on $y$ -axis be $B$ . Let $P = (x, y)$ .

$PA = 3$ cm (from end on $x$ -axis), so $PB = 12 - 3 = 9$ cm.

Let the rod make angle $\theta$ with $x$ -axis.

From $\triangle PBQ$ (where $Q$ is foot of perpendicular from $P$ to $y$ -axis): $\cos\theta = \dfrac{x}{PB} = \dfrac{x}{9}$
From $\triangle PAR$ (where $R$ is foot of perpendicular from $P$ to $x$ -axis): $\sin\theta = \dfrac{y}{PA} = \dfrac{y}{3}$

Using $\cos^2\theta + \sin^2\theta = 1$ :
$\frac{x^2}{81} + \frac{y^2}{9} = 1$

The locus of $P$ is the ellipse $\dfrac{x^2}{81} + \dfrac{y^2}{9} = 1$ .

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6Find the area of the triangle formed by the lines joining the vertex of the parabola

x^2 = 12y

to the ends of its latus rectum.

Given: Parabola $x^2 = 12y$ .

Step 1: Comparing with $x^2 = 4ay$ : $4a = 12 \Rightarrow a = 3$ .

Step 2: Focus $= (0, 3)$ . Ends of latus rectum are at $y = 3$ :
$x^2 = 12(3) = 36 \Rightarrow x = \pm 6$
Ends of latus rectum: $L = (6, 3)$ and $L' = (-6, 3)$ .

Step 3: Vertex $V = (0, 0)$ .

Step 4: Length of latus rectum $= LL' = 12$ .

Height from vertex to latus rectum $=$ distance from $(0,0)$ to line $y = 3$ $= 3$ .

Area $= \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 12 \times 3 = \boxed{18 \text{ sq. units}}$

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7A man running a racecourse notes that the sum of the distances from the two flag posts from him is always

10

m and the distance between the flag posts is

8

m. Find the equation of the posts traced by the man.

Given: Sum of distances from two fixed points $= 10$ m; distance between fixed points $= 8$ m.

Concept: This is the definition of an ellipse where the sum of distances from two foci is constant.

$2a = 10 \Rightarrow a = 5$ ; $2c = 8 \Rightarrow c = 4$ .

$b^2 = a^2 - c^2 = 25 - 16 = 9$

Taking the flag posts on the $x$ -axis:

$\boxed{\frac{x^2}{25} + \frac{y^2}{9} = 1}$

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8An equilateral triangle is inscribed in the parabola

y^2 = 4ax

, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Given: Equilateral triangle inscribed in $y^2 = 4ax$ ; one vertex at origin (vertex of parabola).

Setup: By symmetry, the other two vertices are symmetric about the $x$ -axis. Let one vertex be $P = (at^2, 2at)$ on the parabola.

The side $OP$ makes angle $30°$ with the $x$ -axis (since the triangle is equilateral and symmetric about $x$ -axis, the two sides from origin make angles $\pm 30°$ ).

Step 1: Slope of $OP = \tan 30° = \dfrac{1}{\sqrt{3}}$ .

$\frac{2at}{at^2} = \frac{2}{t} = \frac{1}{\sqrt{3}} \Rightarrow t = 2\sqrt{3}$

Step 2: Coordinates of $P$ :
$x = at^2 = a(12) = 12a, \quad y = 2at = 4a\sqrt{3}$

Step 3: Length of side $OP$ :
$OP = \sqrt{(12a)^2 + (4a\sqrt{3})^2} = \sqrt{144a^2 + 48a^2} = \sqrt{192a^2} = 8a\sqrt{3}$

The length of the side of the equilateral triangle is $8a\sqrt{3}$ .

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Frequently Asked Questions

What are the important topics in Conic Sections for CBSE Class 11 Mathematics?

Key topics in Conic Sections include Conic Sections and Their Classification, Circle, Parabola, Ellipse. Study these first, then practise questions on each for Class 11 exams.

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Exercise 10.1

Exercise 10.2

Exercise 10.3

Exercise 10.4

Miscellaneous Exercise on Chapter 10

Frequently Asked Questions

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