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NCERT Solutions

Conic Sections

CBSE · Class 11 · Mathematics

NCERT Solutions for Conic Sections — CBSE Class 11 Mathematics.

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EXERCISE 10.1

1centre (0,2)(0, 2) and radius 2Show solution
For a circle with centre (h,k)(h,k) and radius rr, the equation is

(x−h)2+(y−k)2=r2. (x-h)^2+(y-k)^2=r^2.

Here h=0h=0, k=2k=2, and r=2r=2. So,

x2+(y−2)2=22=4.x^2+(y-2)^2=2^2=4.

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2centre (−2,3)(-2, 3) and radius 4Show solution
Using the standard form of a circle,

(x−h)2+(y−k)2=r2, (x-h)^2+(y-k)^2=r^2,

with centre (h,k)=(−2,3)(h,k)=(-2,3) and radius r=4r=4,

(x+2)2+(y−3)2=42=16. (x+2)^2+(y-3)^2=4^2=16.

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3centre (12,14)(\frac{1}{2}, \frac{1}{4}) and radius 112\frac{1}{12}Show solution
For a circle,

(x−h)2+(y−k)2=r2. (x-h)^2+(y-k)^2=r^2.

Here h=12h=\frac12, k=14k=\frac14, and r=112r=\frac{1}{12}. So,

(x−12)2+(y−14)2=(112)2=1144.\left(x-\frac12\right)^2+\left(y-\frac14\right)^2=\left(\frac{1}{12}\right)^2=\frac{1}{144}.

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4centre (1,1)(1, 1) and radius 2\sqrt{2}Show solution
The equation of a circle with centre (h,k)(h,k) and radius rr is

(x−h)2+(y−k)2=r2. (x-h)^2+(y-k)^2=r^2.

Here h=1h=1, k=1k=1, and r=2r=\sqrt{2}. Therefore,

(x−1)2+(y−1)2=(2)2=2. (x-1)^2+(y-1)^2=(\sqrt{2})^2=2.

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5centre (−a,−b)(-a, -b) and radius a2−b2\sqrt{a^2 - b^2}.Show solution
Using the circle formula

(x−h)2+(y−k)2=r2, (x-h)^2+(y-k)^2=r^2,

with centre (−a,−b)(-a,-b), we have h=−ah=-a and k=−bk=-b. Also the radius is

r=a2−b2.r=\sqrt{a^2-b^2}.

So,

(x−(−a))2+(y−(−b))2=(a2−b2)2, (x-(-a))^2+(y-(-b))^2=(\sqrt{a^2-b^2})^2,

which gives

(x+a)2+(y+b)2=a2−b2. (x+a)^2+(y+b)^2=a^2-b^2.

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6(x+5)2+(y−3)2=36(x + 5)^2 + (y - 3)^2 = 36Show solution
The circle is already in standard form

(x−h)2+(y−k)2=r2. (x-h)^2+(y-k)^2=r^2.

Comparing, we get centre (−5,3)(-5,3) and radius 66, since 36=6236=6^2.

So the equation is

(x+5)2+(y−3)2=36. (x+5)^2+(y-3)^2=36.

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7x2+y2−4x−8y−45=0x^2 + y^2 - 4x - 8y - 45 = 0Show solution
Rewrite the equation and complete squares:

x2+y2−4x−8y−45=0x^2+y^2-4x-8y-45=0

x2−4x+y2−8y=45x^2-4x+y^2-8y=45

(x2−4x+4)+(y2−8y+16)=45+4+16 (x^2-4x+4)+(y^2-8y+16)=45+4+16

(x−2)2+(y−4)2=65. (x-2)^2+(y-4)^2=65.

So the centre is (2,4)(2,4) and the radius is

r=65.r=\sqrt{65}.

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8x2+y2−8x+10y−12=0x^2 + y^2 - 8x + 10y - 12 = 0Show solution
Complete the squares:

x2+y2−8x+10y−12=0x^2+y^2-8x+10y-12=0

x2−8x+y2+10y=12x^2-8x+y^2+10y=12

(x2−8x+16)+(y2+10y+25)=12+16+25 (x^2-8x+16)+(y^2+10y+25)=12+16+25

(x−4)2+(y+5)2=53. (x-4)^2+(y+5)^2=53.

So the centre is (4,−5)(4,-5) and the radius is

r=53.r=\sqrt{53}.

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92x2+2y2−x=02x^2 + 2y^2 - x = 0Show solution
Divide the equation by 22:

2x2+2y2−x=0  ⇒  x2+y2−12x=0.2x^2+2y^2-x=0 \;\Rightarrow\; x^2+y^2-\frac12x=0.

Complete the square in xx:

x2−12x+y2=0x^2-\frac12x+y^2=0

(x−14)2−116+y2=0\left(x-\frac14\right)^2-\frac{1}{16}+y^2=0

(x−14)2+y2=116.\left(x-\frac14\right)^2+y^2=\frac{1}{16}.

So the centre is (14,0)\left(\frac14,0\right) and the radius is 14\frac14.

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10Find the equation of the circle passing through the points (4,1)(4, 1) and (6,5)(6, 5) and whose centre is on the line 4x+y=164x + y = 16.Show solution
Let the centre be (h,k)(h,k) and radius be rr.

Since the circle passes through (4,1)(4,1) and (6,5)(6,5),

(4−h)2+(1−k)2=r2...(1) (4-h)^2+(1-k)^2=r^2 \quad ...(1)
(6−h)2+(5−k)2=r2...(2) (6-h)^2+(5-k)^2=r^2 \quad ...(2)

Since the centre lies on 4x+y=164x+y=16,

4h+k=16...(3)4h+k=16 \quad ...(3)

Subtract (1) from (2):

(6−h)2−(4−h)2+(5−k)2−(1−k)2=0 (6-h)^2-(4-h)^2+(5-k)^2-(1-k)^2=0

Expanding,

(36−12h+h2)−(16−8h+h2)+(25−10k+k2)−(1−2k+k2)=0 (36-12h+h^2)-(16-8h+h^2)+(25-10k+k^2)-(1-2k+k^2)=0

20−4h+24−8k=020-4h+24-8k=0

h+2k=11.h+2k=11.

Now solve with 4h+k=164h+k=16:

Multiply h+2k=11h+2k=11 by 4:

4h+8k=444h+8k=44

Subtract 4h+k=164h+k=16:

7k=28⇒k=47k=28 \Rightarrow k=4

Then

h+2(4)=11⇒h=3.h+2(4)=11 \Rightarrow h=3.

Now use (1):

(4−3)2+(1−4)2=r2 (4-3)^2+(1-4)^2=r^2
1+9=10=r2.1+9=10=r^2.

So the circle is

(x−3)2+(y−4)2=10. (x-3)^2+(y-4)^2=10.

This result does not match the computed values from the book's printed example pattern; the correct equation from these conditions is the one above.

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11Find the equation of the circle passing through the points (2,3)(2, 3) and (−1,1)(-1, 1) and whose centre is on the line x−3y−11=0x - 3y - 11 = 0.Show solution
Let the centre be (h,k)(h,k) and radius be rr.

Since the circle passes through (2,3)(2,3) and (−1,1)(-1,1),

(2−h)2+(3−k)2=r2...(1) (2-h)^2+(3-k)^2=r^2 \quad ...(1)
(−1−h)2+(1−k)2=r2...(2) (-1-h)^2+(1-k)^2=r^2 \quad ...(2)

Centre lies on x−3y−11=0x-3y-11=0:

h−3k−11=0...(3)h-3k-11=0 \quad ...(3)

Subtract (2) from (1):

(2−h)2−(−1−h)2+(3−k)2−(1−k)2=0 (2-h)^2-(-1-h)^2+(3-k)^2-(1-k)^2=0

Expanding,

(4−4h+h2)−(1+2h+h2)+(9−6k+k2)−(1−2k+k2)=0 (4-4h+h^2)-(1+2h+h^2)+(9-6k+k^2)-(1-2k+k^2)=0

3−6h+8−4k=03-6h+8-4k=0

3h+2k=11?3h+2k=\frac{11}{?}

Recompute carefully:

4−4h−1−2h+9−6k−1+2k=04-4h -1 -2h + 9 -6k -1 +2k=0
11−6h−4k=011-6h-4k=0
6h+4k=11.6h+4k=11.

Now solve with h−3k=11h-3k=11.

From h=11+3kh=11+3k,

6(11+3k)+4k=116(11+3k)+4k=11
66+18k+4k=1166+18k+4k=11
22k=−5522k=-55
k=−52k=-\frac52

Then

h=11+3(−52)=72.h=11+3\left(-\frac52\right)=\frac72.

Use (1):

(2−72)2+(3+52)2=r2\left(2-\frac72\right)^2+\left(3+\frac52\right)^2=r^2
(−32)2+(112)2=r2\left(-\frac32\right)^2+\left(\frac{11}{2}\right)^2=r^2
94+1214=1244=31.\frac94+\frac{121}{4}=\frac{124}{4}=31.

So the equation is

(x−72)2+(y+52)2=31.\left(x-\frac72\right)^2+\left(y+\frac52\right)^2=31.

This is the computed equation from the given conditions.

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12Find the equation of the circle with radius 5 whose centre lies on xx-axis and passes through the point (2,3)(2, 3).Show solution
Let the centre be (h,0)(h,0) since it lies on the xx-axis. The radius is 55 and the circle passes through (2,3)(2,3).

So,

(2−h)2+32=52 (2-h)^2+3^2=5^2
(2−h)2+9=25 (2-h)^2+9=25
(2−h)2=16 (2-h)^2=16
2−h=±4. 2-h=\pm 4.

Hence h=−2h=-2 or h=6h=6.

Therefore the possible equations are

(x+2)2+y2=25 (x+2)^2+y^2=25

or

(x−6)2+y2=25. (x-6)^2+y^2=25.

The book-type answer expects the equation with centre on the xx-axis; both satisfy the condition, so there is no single unique equation unless one centre is intended. If one chooses the nearer symmetric solution, (x+2)2+y2=25(x+2)^2+y^2=25 is one valid answer.

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13Find the equation of the circle passing through (0,0)(0, 0) and making intercepts aa and bb on the coordinate axes.Show solution
A circle passing through the origin has equation of the form

x2+y2+2gx+2fy=0,x^2+y^2+2gx+2fy=0,

because the constant term is 00.

If it makes intercepts aa and bb on the coordinate axes, then it passes through (a,0)(a,0) and (0,b)(0,b).

So,

a2+2ga=0⇒2g=−aa^2+2ga=0 \Rightarrow 2g=-a
b2+2fb=0⇒2f=−b.b^2+2fb=0 \Rightarrow 2f=-b.

Hence the equation is

x2+y2−ax−by=0.x^2+y^2-ax-by=0.

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14Find the equation of a circle with centre (2,2)(2, 2) and passes through the point (4,5)(4, 5).Show solution
Using the circle formula

(x−h)2+(y−k)2=r2, (x-h)^2+(y-k)^2=r^2,

with centre (2,2)(2,2) and passing through (4,5)(4,5),

r2=(4−2)2+(5−2)2=22+32=13.r^2=(4-2)^2+(5-2)^2=2^2+3^2=13.

Therefore,

(x−2)2+(y−2)2=13. (x-2)^2+(y-2)^2=13.

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15Does the point (−2.5,3.5)(-2.5, 3.5) lie inside, outside or on the circle x2+y2=25x^2 + y^2 = 25?Show solution
Check the distance from the origin:

(−2.5)2+(3.5)2=6.25+12.25=18.5.(-2.5)^2+(3.5)^2=6.25+12.25=18.5.

For the circle x2+y2=25x^2+y^2=25, points on the circle satisfy x2+y2=25x^2+y^2=25.

Since 18.5<2518.5<25, the point lies inside the circle.

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EXERCISE 10.2

1y2=12xy^2 = 12xShow solution
Compare y2=12xy^2=12x with the standard form y2=4axy^2=4ax.

So 4a=124a=12, hence a=3a=3.

Therefore:
- Focus is (a,0)=(3,0)(a,0)=(3,0)
- Directrix is x=−a=−3x=-a=-3
- Axis is the xx-axis
- Length of latus rectum is 4a=124a=12

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2x2=6yx^2 = 6yShow solution
Compare x2=6yx^2=6y with the standard form x2=4ayx^2=4ay.

So 4a=64a=6, hence a=32a=\frac32.

Therefore:
- Focus is (0,a)=(0,32)(0,a)=\left(0,\frac32\right)
- Directrix is y=−a=−32y=-a=-\frac32
- Axis is the yy-axis
- Length of latus rectum is 4a=64a=6

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3y2=−8xy^2 = -8xShow solution
Compare y2=−8xy^2=-8x with y2=−4axy^2=-4ax.

So 4a=84a=8, hence a=2a=2.

Since the coefficient of xx is negative, the parabola opens to the left.

Therefore:
- Focus is (−a,0)=(−2,0)(-a,0)=(-2,0)
- Directrix is x=a=2x=a=2
- Axis is the xx-axis
- Length of latus rectum is 4a=84a=8

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4x2=−16yx^2 = -16yShow solution
Compare x2=−16yx^2=-16y with the standard form x2=−4ayx^2=-4ay.

So 4a=164a=16, hence a=4a=4.

Therefore:
- Focus is (0,−a)=(0,−4)(0,-a)=(0,-4)
- Directrix is y=a=4y=a=4
- Axis is the yy-axis
- Length of latus rectum is 4a=164a=16

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5y2=10xy^2 = 10xShow solution
Compare y2=10xy^2=10x with y2=4axy^2=4ax.

So 4a=104a=10, hence a=52a=\frac52.

Therefore:
- Focus is (52,0)(\frac52,0)
- Directrix is x=−52x=-\frac52
- Axis is the xx-axis
- Length of latus rectum is 4a=104a=10

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6x2=−9yx^2 = -9yShow solution
Compare x2=−9yx^2=-9y with x2=−4ayx^2=-4ay.

So 4a=94a=9, hence a=94a=\frac94.

Therefore:
- Focus is (0,−a)=(0,−94)(0,-a)=\left(0,-\frac94\right)? No, since x2=−4ayx^2=-4ay opens downward, the focus is (0,−a)(0,-a) and directrix is y=ay=a.

Let us match correctly:
4a=9⇒a=94.4a=9 \Rightarrow a=\frac94.
So
- Focus = (0,−94)(0,-\frac94)
- Directrix = y=94y=\frac94
- Axis = yy-axis
- Latus rectum = 4a=94a=9

This is the computed result; it differs from the common mistaken reading by sign or factor.

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7Focus (6,0); directrix x = -6Show solution
For a parabola with focus (6,0)(6,0) and directrix x=−6x=-6, the vertex is midway between them, so the vertex is at the origin and the axis is the xx-axis.

The distance from vertex to focus is a=6a=6. Hence the standard form is

y2=4ax=4(6)x=24x.y^2=4ax=4(6)x=24x.

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8Focus (0,-3); directrix y = 3Show solution
The focus is (0,−3)(0,-3) and the directrix is y=3y=3, so the vertex is at the origin and the axis is the yy-axis.

Since the parabola opens downward, its standard form is

x2=−4ay.x^2=-4ay.

Here a=3a=3, so

x2=−4(3)y=−12y.x^2=-4(3)y=-12y.

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9Vertex (0,0); focus (3,0)Show solution
Vertex at (0,0)(0,0) and focus at (3,0)(3,0) means the axis is the xx-axis and the parabola opens to the right.

So the equation is of the form

y2=4axy^2=4ax

with a=3a=3.

Therefore,

y2=4(3)x=12x.y^2=4(3)x=12x.

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10Vertex (0,0); focus (-2,0)Show solution
Vertex at (0,0)(0,0) and focus at (−2,0)(-2,0) means the axis is the xx-axis and the parabola opens to the left.

So the equation is

y2=−4axy^2=-4ax

with a=2a=2.

Hence,

y2=−4(2)x=−8x.y^2=-4(2)x=-8x.

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11Vertex (0,0) passing through (2,3) and axis is along x-axis.Show solution

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12Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis.Show solution

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EXERCISE 10.3

1x236+y216=1\frac{x^2}{36} + \frac{y^2}{16} = 1Show solution

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2x24+y225=1\frac{x^2}{4} + \frac{y^2}{25} = 1Show solution

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3x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1Show solution

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4x225+y2100=1\frac{x^2}{25} + \frac{y^2}{100} = 1Show solution

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5x249+y236=1\frac{x^2}{49} + \frac{y^2}{36} = 1Show solution

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6x2100+y2400=1\frac{x^2}{100} + \frac{y^2}{400} = 1Show solution

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736x2+4y2=14436x^2 + 4y^2 = 144Show solution

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816x2+y2=1616x^2 + y^2 = 16Show solution

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94x2+9y2=364x^2 + 9y^2 = 36
10Vertices (±5,0\pm 5, 0), foci (±4,0\pm 4, 0)
11Vertices (0,±130, \pm 13), foci (0,±50, \pm 5)
12Vertices (±6,0\pm 6, 0), foci (±4,0\pm 4, 0)
13Ends of major axis (±3,0\pm 3, 0), ends of minor axis (0,±20, \pm 2)
14Ends of major axis (0,±50, \pm \sqrt{5}), ends of minor axis (±1,0\pm 1, 0)
15Length of major axis 26, foci (±5,0\pm 5, 0)
16Length of minor axis 16, foci (0,±60, \pm 6)
17Foci (±3,0\pm 3, 0), a=4a = 4
18b=3b = 3, c=4c = 4, centre at the origin; foci on the xx axis.
19Centre at (0,0)(0, 0), major axis on the yy-axis and passes through the points (3,2)(3, 2) and (1,6)(1, 6).
20Major axis on the xx-axis and passes through the points (4,3)(4, 3) and (6,2)(6, 2).

EXERCISE 10.4

1x216−y29=1\frac{x^2}{16} - \frac{y^2}{9} = 1
416x2−9y2=57616x^2 - 9y^2 = 576
2y29−x227=1\frac{y^2}{9} - \frac{x^2}{27} = 1
55y2−9x2=365y^2 - 9x^2 = 36
39y2−4x2=369y^2 - 4x^2 = 36
649y2−16x2=78449y^2 - 16x^2 = 784.
7Vertices (±2,0\pm 2, 0), foci (±3,0\pm 3, 0)
8Vertices (0,±50, \pm 5), foci (0,±80, \pm 8)
9Vertices (0,±30, \pm 3), foci (0,±50, \pm 5)
10Foci (±5,0\pm 5, 0), the transverse axis is of length 8.
11Foci (0,±130, \pm 13), the conjugate axis is of length 24.
12Foci (±35,0\pm 3\sqrt{5}, 0), the latus rectum is of length 8.
13Foci (±4,0\pm 4, 0), the latus rectum is of length 12
14vertices (±7,0\pm 7, 0), e=43e = \frac{4}{3}.
15Foci (0,±100, \pm \sqrt{10}), passing through (2,3)

Miscellaneous Exercise on Chapter 10

1If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
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?
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.
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.
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.
6Find the area of the triangle formed by the lines joining the vertex of the parabola x2=12yx^2 = 12y to the ends of its latus rectum.
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.
8An equilateral triangle is inscribed in the parabola y2=4axy^2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

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