Conic Sections

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

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

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

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

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

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 $= 6$.

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 $= \sqrt{65}$.

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 $= \sqrt{53}$.

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 $= \dfrac{1}{4}$.

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

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

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., $x^2 + y^2 - 12x + 11 = 0$

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

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

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

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.

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: $4a = 12$

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: $4a = 6$

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: $4a = 8$

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: $4a = 16$

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: $4a = 10$

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: $4a = 9$

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 $a = 6$.

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

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 $a = 3$.

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

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

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

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

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

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

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

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

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

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

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

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 16}{6} = \dfrac{16}{3}$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{5} = \dfrac{8}{5}$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 9}{4} = \dfrac{9}{2}$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 25}{10} = 5$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 36}{7} = \dfrac{72}{7}$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 100}{20} = 10$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{6} = \dfrac{4}{3}$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 1}{4} = \dfrac{1}{2}$

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: $\dfrac{2b^2}{a} = \dfrac{2 \times 4}{3} = \dfrac{8}{3}$

Given: Vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$.

Since vertices and foci are on $x$-axis: $a = 5$, $c = 4$.
$$b^2 = a^2 - c^2 = 25 - 16 = 9$$

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

Given: Vertices $(0, \pm 13)$, foci $(0, \pm 5)$.

Major axis along $y$-axis: $a = 13$, $c = 5$.
$$b^2 = a^2 - c^2 = 169 - 25 = 144$$

$$\boxed{\frac{x^2}{144} + \frac{y^2}{169} = 1}$$

Given: Vertices $(\pm 6, 0)$, foci $(\pm 4, 0)$.

Major axis along $x$-axis: $a = 6$, $c = 4$.
$$b^2 = 36 - 16 = 20$$

$$\boxed{\frac{x^2}{36} + \frac{y^2}{20} = 1}$$