Analytical Geometry - Conics - Ellipse

Find the equation of ellipse whose latus-rectum is $5$ and eccentricity is $\dfrac{2}{3}$.

Latus-rectum of ellipse $= \dfrac{2b^2}{a} = 5$ (given)

$\implies$ $b^2 = \dfrac{5 a}{2}$ $\;\;\; \cdots \; (1)$

Eccentricity $= e = \dfrac{2}{3}$ (given)

For an ellipse, $\;$ $b^2 = a^2 \left(1 - e^2\right)$ $\;\;\; \cdots \; (2)$

Substituting the values of $b^2$ and $e$ in equation $(2)$ gives,

$\dfrac{5a}{2} = a^2 \left[1 - \left(\dfrac{2}{3}\right)^2\right]$

i.e. $\;$ $\dfrac{5}{2} = \dfrac{5a}{9}$

i.e. $\;$ $a = \dfrac{9}{2}$ $\implies$ $a^2 = \dfrac{81}{4}$

Substituting the value of $a$ in equation $(1)$ gives,

$b^2 = 5 \times \dfrac{9}{2} \times \dfrac{1}{2} = \dfrac{45}{4}$

The equation of ellipse is $\;\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

i.e. $\;$ $\dfrac{x^2}{81 / 4} + \dfrac{y^2}{45 / 4} = 1$

i.e. $\;$ $\dfrac{4 x^2}{81} + \dfrac{4 y^2}{45} = 1$

Analytical Geometry - Conics - Ellipse

The orbit of the planet mercury around the sun is in an elliptical shape with the sun at a focus. The semi-major axis is of length 36 million miles and the eccentricity of the orbit is 0.206. Find

1. how close does mercury get to the sun;
2. the greatest possible distance between mercury and sun.

Let the equation of the elliptical orbit be: $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

Center of the ellipse is at $C \left(0,0\right)$.

Let the sun be at focus $F_1$.

Vertices of the ellipse are $A_1$ and $A_2$.

Semi-major axis of the ellipse $= CA_1 = CA_2 = a = 36$ million miles

Eccentricity of the elliptical orbit $= e = 0.206$

Now, $CF_1 = a \times e = 36 \times 0.206 = 7.416$ million miles

1. When mercury is at point $A_1$, it is closest to the sun.
   
   $\therefore$ $\;$ Required distance $= F_1 A_1$
   
   Now, $F_1 A_1 = CA_1 - CF_1 = 36 - 7.416 = 28.584$ million miles
   
   
2. When mercury is at the point $A_2$, it is farthest from the sun.
   
   $\therefore$ $\;$ Required distance $= F_1 A_2$
   
   Now, $F_1 A_2 = CF_1 + CA_2 = 7.416 + 36 = 43.416$ million miles

Analytical Geometry - Conics - Ellipse

Find the eccentricity, center, foci and vertices of the ellipse $\;$ $9x^2 + 4y^2 = 36$ $\;$ and sketch it.

Equation of given ellipse is: $\;$ $9x^2 + 4y^2 = 36$

i.e. $\;$ $\dfrac{x^2}{36/9} + \dfrac{y^2}{36/4} = 1$

i.e. $\;$ $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$ $\;\;\; \cdots \; (1)$

Major axis of the ellipse is along the Y axis.

Comparing equation $(1)$ with the standard equation of the ellipse $\;$ $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ $\;$ gives

$b^2 = 4 \implies b = 2$; $\;$ and $\;$ $a^2 = 9 \implies a = 3$

Now, $e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{4}{9}} = \dfrac{\sqrt{5}}{3}$

$\therefore$ $\;$ $a \times e = 3 \times \dfrac{\sqrt{5}}{3} = \sqrt{5}$

Center $= \left(0, 0\right)$

Foci $\left(F_1 \text{ and } F_2\right) = \left(0, \pm ae\right) = \left(0, \pm \sqrt{5}\right)$

Vertices $\left(A_1 \text{ and } A_2\right) = \left(0, \pm a\right) = \left(0, \pm 3\right)$

Analytical Geometry - Conics - Ellipse

Find the eccentricity, center, foci and vertices of the ellipse $\;$ $x^2 + 4y^2 -8x - 16 y - 68 = 0$ $\;$ and sketch it.

Equation of the given ellipse is: $\;$ $x^2 + 4y^2 - 8x - 16 y - 68 = 0$

i.e. $\;$ $\left(x^2 - 8x\right) + 4 \left(y^2 - 4y\right) = 68$

i.e. $\;$ $\left[\left(x^2 - 8x + 16\right) - 16\right] + 4 \left[\left(y^2 - 4y + 4\right) - 4\right] = 68$

i.e. $\;$ $\left(x - 4\right)^2 + 4 \left(y - 2\right)^2 = 68 + 16 + 16$

i.e. $\;$ $\left(x - 4\right)^2 + 4 \left(y - 2\right)^2 = 100$

i.e. $\;$ $\dfrac{\left(x - 4\right)^2}{100} + \dfrac{\left(y - 2\right)^2}{100 / 4} = 1$

i.e. $\;$ $\dfrac{\left(x - 4\right)^2}{100} + \dfrac{\left(y - 2\right)^2}{25} = 1$ $\;\;\; \cdots \; (1)$

Let $X = x - 4$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 2$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ can be written as: $\;$ $\dfrac{X^2}{100} + \dfrac{Y^2}{25} = 1$ $\;\;\; \cdots \; (3)$

Major axis of the ellipse given by equation $(3)$ is along the X axis.

Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{a^2} + \dfrac{Y^2}{b^2} = 1$ $\;$ gives

$a^2 = 100$ $\implies$ $a = 10$; $\;\;$ $b^2 = 25$ $\implies$ $b = 5$

$\therefore$ $\;$ Eccentricity $= e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{25}{100}} = \sqrt{\dfrac{75}{100}} = \dfrac{\sqrt{3}}{2}$

Now, $a \times e = 10 \times \dfrac{\sqrt{3}}{2} = 5 \sqrt{3}$

	Referred to X, Y	Referred to x, y $X = x - 4$; $\;$ $Y = y - 2$ i.e. $\;$ $x = X + 4$; $\;$ $y = Y + 2$
Center	$\left(0, 0\right)$	$X = 0 \implies x = 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ Center $= \left(4, 2\right)$
Foci	$F_1 = \left(ae, 0\right)$ $F_1 = \left(10 \times \dfrac{\sqrt{3}}{2}, 0\right)$ i.e. $\;$ $F_1 = \left(5 \sqrt{3}, 0\right)$	$X = 5 \sqrt{3} \implies x = 5 \sqrt{3} + 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $F'_1 = \left(4 + 5 \sqrt{3}, 2\right)$
	$F_2 = \left(-ae, 0\right)$ i.e. $\;$ $F_2 = \left(-10 \times \dfrac{\sqrt{3}}{2}, 0\right)$ i.e. $\;$ $F_2 = \left(-5 \sqrt{3}, 0\right)$	$X = -5 \sqrt{3} \implies x = -5 \sqrt{3} + 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $F'_2 = \left(4 - 5 \sqrt{3}, 2\right)$
Vertices	$A_1 = \left(+a, 0\right)$ i.e. $\;$ $A_1 = \left(10, 0\right)$	$X = 10 \implies x = 14$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $A'_1 = \left(14, 2\right)$
	$A_2 = \left(-a, 0\right)$ i.e. $\;$ $A_2 = \left(-10, 0\right)$	$X = -10 \implies x = -6$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $A'_2 = \left(-6, 2\right)$

Analytical Geometry - Conics - Ellipse

Find the equations of directrices, latus rectum and length of latus rectum for the ellipse $3x^2 + 2y^2 - 30 x - 4y + 23 = 0$

Equation of given ellipse is: $\;$ $3x^2 + 2y^2 - 30 x - 4y + 23 = 0$

i.e. $\;$ $3 \left(x^2 - 10x\right) + 2 \left(y^2 - 2y\right) = -23$

i.e. $\;$ $3 \left[\left(x^2 - 10x + 25\right) - 25\right] + 2 \left[\left(y^2 - 2y + 1\right) -1\right] = -23$

i.e. $\;$ $3 \left(x - 5\right)^2 + 2 \left(y - 1\right)^2 - 75 - 2 = - 23$

i.e. $\;$ $3 \left(x - 5\right)^2 + 2 \left(y - 1\right)^2 = 54$

i.e. $\;$ $\dfrac{\left(x - 5\right)^2}{54 / 3} + \dfrac{\left(y - 1\right)^2}{54 / 2} = 1$

i.e. $\;$ $\dfrac{\left(x - 5\right)^2}{18} + \dfrac{\left(y - 1\right)^2}{27} = 1$ $\;\;\; \cdots \; (1)$

Let $\;$ $X = x - 5$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 1$ $\;\;\; \cdots \; (2b)$

Then in view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$\dfrac{X^2}{18} + \dfrac{Y^2}{27} = 1$ $\;\;\; \cdots \; (3)$

Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{b^2} + \dfrac{Y^2}{a^2} = 1$ $\;$ (for $b < a$) $\;$ gives

$b^2 = 18$ $\;\;\; \cdots \; (4a)$; $\;$ $a^2 = 27 \implies a = 3 \sqrt{3}$ $\;\;\; \cdots \; (4b)$

Now, $e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{18}{27} = \dfrac{9}{27}$ $\implies$ $e = \dfrac{1}{\sqrt{3}}$ $\;\;\; \cdots \; (4c)$

Equations of directrices are $\;$ $Y = \pm \dfrac{a}{e}$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equations $(2b)$, $(4b)$ and $(4c)$, equation $(5)$ becomes

$y - 1 = \pm \dfrac{3 \sqrt{3}}{3}$

i.e. $\;$ $y - 1 = \pm 9$

i.e. $\;$ $y = 10$ $\;\;\; \cdots \; (5a)$ $\;$ and $\;$ $y = -8$ $\;\;\; \cdots \; (5b)$

Equations of latus rectum are: $\;$ $Y = \pm ae$ $\;\;\; \cdots \; (6)$

$\therefore$ $\;$ In view of equations $(2b)$, $(4b)$ and $(4c)$, equation $(6)$ becomes

$y - 1 = \pm 3 \sqrt{3 \times \dfrac{1}{\sqrt{3}}}$

i.e. $\;$ $y - 1 = \pm 3$

i.e. $\;$ $y = 4$ $\;\;\; \cdots \; (6a)$ $\;$ and $y = -2$ $\;\;\; \cdots \; (6b)$

Length of latus rectum $= \dfrac{2b^2}{a} = \dfrac{2 \times 18}{3 \sqrt{3}} = 4 \sqrt{3}$ $\;$ [from equations $(4a)$ and $(4b)$]

Analytical Geometry - Conics - Ellipse

Find the equations of directrices, latus rectum and length of latus rectum for the ellipse $9x^2 + 16y^2 = 144$

Equation of the given ellipse is: $\;\;$ $9x^2 + 16y^2 = 144$

i.e. $\;$ $\dfrac{x^2}{144 / 9} + \dfrac{y^2}{144 / 16} = 1$

i.e. $\;$ $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 16 \implies a = 4$; $\;$ $b^2 = 9$

Now, $\;$ $e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{9}{16} = \dfrac{7}{16}$ $\implies$ $e = \dfrac{\sqrt{7}}{4}$

Equations of directrices are: $\;$ $x = \pm \dfrac{a}{e}$

i.e. $\;$ $x = \pm \dfrac{4}{\sqrt{7} / 4}$

i.e. $\;$ $x = \pm \dfrac{16}{\sqrt{7}}$

Equations of latus rectum are: $\;$ $x = \pm ae$

i.e. $\;$ $x = \pm 4 \times \dfrac{\sqrt{7}}{4}$

i.e. $\;$ $x = \pm \sqrt{7}$

Length of latus rectum $= \dfrac{2b^2}{a} = 2 \times \dfrac{9}{4} = \dfrac{9}{2}$

Analytical Geometry - Conics - Ellipse

Find the equation and length of major and minor axes of $\;$ $9x^2 + 4y^2 = 20$

Equation of the given ellipse is: $\;$ $9x^2 + 4y^2 = 20$

i.e. $\;$ $\dfrac{x^2}{20 / 9} + \dfrac{y^2}{20 / 4} =1$

i.e. $\;$ $\dfrac{x^2}{20 / 9} + \dfrac{y^2}{5} = 1$ $\;\;\; \cdots \; (1)$

The major axis of the given ellipse is along the Y axis and the minor axis is along the X axis.

$\therefore$ $\;$ Equation of major axis is: $\;$ $x = 0$

Equation of minor axis is: $\;$ $y = 0$

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ $\;$ $\text{for } \left(b < a\right)$ $\;$ gives

$a^2 = 5 \implies a = \sqrt 5$ $\;$ and $\;$ $b^2 = \dfrac{20}{9} \implies b = \dfrac{2 \sqrt{5}}{3}$

$\therefore$ $\;$ Length of major axis $= 2a = 2 \sqrt{5}$

Length of minor axis $= 2b = 2 \times \dfrac{2 \sqrt{5}}{3} = \dfrac{4 \sqrt{5}}{3}$

Analytical Geometry - Conics - Ellipse

Find the equation and length of major and minor axes of $\;$ $5x^2 + 9y^2 + 10 x - 36 y - 4 = 0$

Equation of the given ellipse is: $\;$ $5x^2 + 9y^2 + 10 x - 36 y - 4 = 0$

i.e. $\;$ $\left(5x^2 + 10x\right) + \left(9y^2 - 36y\right) = 4$

i.e. $\;$ $5 \left(x^2 + 2x\right) + 9 \left(y^2 - 4y\right) = 4$

i.e. $\;$ $5 \left[\left(x^2 + 2x + 1\right) - 1\right] + 9 \left[\left(y^2 - 4y + 4\right) -4\right] = 4$

i.e. $\;$ $5 \left[\left(x + 1\right)^2 - 1\right] + 9 \left[\left(y - 2\right)^2 - 4\right] = 4$

i.e. $\;$ $5 \left(x + 1\right)^2 + 9 \left(y - 2\right)^2 = 4 + 5 + 36$

i.e. $\;$ $5 \left(x + 1\right)^2 + 9 \left(y - 2\right)^2 = 45$

i.e. $\;$ $\dfrac{\left(x + 1\right)^2}{45 / 5} + \dfrac{\left(y - 2\right)^2}{45 / 9} = 1$

i.e. $\;$ $\dfrac{\left(x + 1\right)^2}{9} + \dfrac{\left(y -2 \right)^2}{5} = 1$ $\;\;\; \cdots \; (1)$

Let $\;$ $X = x + 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 2$ $\;\;\; \cdots \; (2b)$

Then, in view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$\dfrac{X^2}{9} + \dfrac{Y^2}{5} = 1$ $\;\;\; \cdots \; (3)$

The major axis is along the X axis and the minor axis is along the Y axis.

$\therefore$ $\;$ Equation of major axis is: $\;$ $Y = 0$ $\implies$ $y - 2 = 0$

Equation of minor axis is: $\;$ $X = 0$ $\implies$ $x + 1 = 0$

Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{a^2} + \dfrac{Y^2}{b^2} = 1$ $\;$ $\left(\text{for } a > b\right)$ $\;$ gives

$a^2 = 9 \implies a = 3$; $\;$ and $\;$ $b^2 = 5 \implies b = \sqrt{5}$

$\therefore$ $\;$ Length of major axis $= 2a = 2 \times 3 = 6$

Length of minor axis $= 2b = 2 \sqrt{5}$

Analytical Geometry - Conics - Ellipse

Find the locus of a point which moves so that the sum of its distances from $\left(3,0\right)$ and $\left(-3,0\right)$ is $9$.

Let $P \left(x, y\right)$ be the variable point.

Let $F_1 = \left(3,0\right)$ and $F_2 = \left(-3,0\right)$

Then as per question $\;$ $PF_1 + PF_2 = 9$

i.e. $\;$ $\sqrt{\left(x - 3\right)^2 + \left(y - 0\right)^2} + \sqrt{\left(x + 3\right)^2 + \left(y - 0\right)^2} = 9$

i.e. $\;$ $\sqrt{x^2 - 6x + 9 + y^2} - 9 = \sqrt{x^2 + 6x + 9 + y^2}$ $\;\;\; \cdots \; (1)$

Squaring both sides of equation $(1)$ gives

$x^2 - 6x + 9 + y^2 + 81 - 18 \sqrt{x^2 - 6x + 9 + y^2} = x^2 + 6 x + 9 + y^2$

i.e. $\;$ $- 12 x + 81 = 18 \sqrt{x^2 - 6x + 9 + y^2}$

i.e. $\;$ $-4 x + 27 = 6 \sqrt{x^2 - 6x + 9 + y^2}$ $\;\;\; \cdots \; (2)$

Squaring both sides of equation $(2)$ gives

$16x^2 + 729 - 216 x = 36 \left(x^2 - 6x + 9 + y^2\right)$

i.e. $\;$ $20 x^2 + 36 y^2 = 405$

i.e. $\;$ $\dfrac{x^2}{405 / 20} + \dfrac{y^2}{405 / 36} = 1$

i.e. $\;$ $\dfrac{x^2}{81 / 4} + \dfrac{y^2}{45 / 4} = 1$ $\;\;\; \cdots \; (3)$

Equation $(3)$ gives the required equation of locus.

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if the foci are $\left(\pm 3, 0\right)$ and the length of the latus rectum is $\dfrac{32}{5}$

Let the foci be $\;$ $F_1 = \left(3,0\right)$; $\;\;$ $F_2 = \left(-3,0\right)$

Since the foci lie on the X axis, the major axis of the ellipse is along the X axis.

$\therefore$ $\;$ Let the equation of the required ellipse be

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;\;\; \cdots \; (1)$

Now, length$\left(F_1 F_2\right) = 2ae$ $\;\;\; \cdots \; (2)$

length of latus rectum $= \dfrac{2b^2}{a}$ $\;\;\; \cdots \; (3)$

and $\;$ $b^2 = a^2 \left(1 - e^2\right)$ $\;\;\; \cdots \; (4)$

Given: $\;\;$ Length of latus rectum $= \dfrac{32}{5}$

$\implies$ $\dfrac{2b^2}{a} = \dfrac{32}{5}$ $\implies$ $b^2 = \dfrac{16a}{5}$ $\;\;\; \cdots \; (5a)$

Also, $\;$ $F_1 F_2 = \sqrt{\left(3 + 3\right)^2} = 6$

$\therefore$ $\;$ We have $\;\;$ $6 = 2ae$ $\implies$ $e = \dfrac{3}{a}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ We have from equations $(4)$, $(5a)$ and $(5b)$,

$\dfrac{16a}{5} = a^2 \left(1 - \dfrac{9}{a^2}\right)$

i.e. $\;$ $\dfrac{16 a}{5} = a^2 - 9$

i.e. $\;$ $5a^2 - 16a - 45 = 0$

i.e. $\;$ $\left(a - 5\right) \left(5a + 9\right) = 0$

i.e. $\;$ $a = 5$ $\;\;\; \cdots \; (6)$ $\;$ or $\;$ $a = \dfrac{-9}{5}$ $\;$ (not an acceptable solution)

$\therefore$ $\;$ In view of equation $(6)$, we have from equation $(5a)$

$b^2 = \dfrac{16 \times 5}{5} = 16$

$\therefore$ $\;$ Substituting the values of $a$ and $b^2$ in equation $(1)$ gives the equation of the required ellipse as

$\dfrac{x^2}{5^2} + \dfrac{y^2}{16} = 1$

i.e. $\;$ $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if the center is $\left(3, -4\right)$, one of the foci is $\left(3 + \sqrt{3}, - 4\right)$ and $e = \dfrac{\sqrt{3}}{2}$

Let the center of the ellipse be $= C \left(h, k\right) = \left(3, -4\right)$

Let one of the foci be $= F_1 = \left(3 + \sqrt{3}, -4\right)$

Let the other focus be $= F_2 = \left(x_2, y_2\right)$

Now, center of the ellipse is the midpoint of $F_1 F_2$.

$\therefore$ $\;$ We have $\;$ $3 = \dfrac{3 + \sqrt{3} + x_2}{2}$; $\;$ $-4 = \dfrac{-4 + y_2}{2}$

i.e. $\;$ $x_2 = 3 - \sqrt{3}$; $\;$ $y_2 = -4$

$\therefore$ $\;$ $F_2 = \left(x_2, y_2\right) = \left(3 - \sqrt{3}, -4\right)$

$\therefore$ $\;$ From the given data, the major axis is parallel to the X axis.

$\therefore$ $\;$ Let the equation of the required ellipse be: $\;$ $\dfrac{\left(x - h\right)^2}{a^2} + \dfrac{\left(y - k\right)^2}{b^2} = 1$

Given: $\;$ $e = \dfrac{\sqrt{3}}{2}$

Now, $\;$ $F_1 F_2 = 2 a e$

i.e. $\;$ $\sqrt{\left(3 + \sqrt{3} - 3 + \sqrt{3}\right)^2 + \left(-4 + 4\right)^2} = 2 \times a \times \dfrac{\sqrt{3}}{2}$

i.e. $\;$ $2 \sqrt{3} = a \sqrt{3}$ $\implies$ $a = 2$

Further, $\;$ $b^2 = a^2 \left(1 - e^2\right)$

i.e. $\;$ $b^2 = \left(2\right)^2 \times \left[1 - \left(\dfrac{\sqrt{3}}{2}\right)^2\right]$

i.e. $\;$ $b^2 = 4 \times \dfrac{1}{4}$ $\implies$ $b^2 = 1$

$\therefore$ $\;$ The equation of the required ellipse is

$\dfrac{\left(x - 3\right)^2}{2^2} + \dfrac{\left(y + 4\right)^2}{2} = 1$

i.e. $\;$ $\dfrac{\left(x - 3\right)^2}{4} + \dfrac{\left(y + 4\right)^2}{1} = 1$

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if the foci are $\left(\pm 3, 0\right)$ and the vertices are $\left(\pm 5, 0\right)$.

Let the foci be $F_1 = \left(3,0\right)$ and $F_2 = \left(-3,0\right)$

$\therefore$ $\;$ The foci $F_1$ and $F_2$ lie on the the X axis.

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

Let the vertices be $V_1 = \left(5,0\right)$ and $V_2 = \left(-5,0\right)$

Now, $\text{length } \left(V_1 V_2\right) = 2a$

i.e. $\;$ $\sqrt{\left(5 + 5\right)^2} = 2a$

i.e. $\;$ $10 = 2a$ $\implies$ $a = 5$

and $\text{length } \left(F_1 F_2\right) = 2 a e$

i.e. $\;$ $\sqrt{\left(3 + 3\right)^2} = 2 \times 5 \times e$

i.e. $\;$ $6 = 10 e$ $\implies$ $e = \dfrac{3}{5}$

Also, $\;$ $b^2 = a^2 \left(1 - e^2\right)$

i.e. $\;$ $b^2 = \left(5\right)^2 \times \left[1 - \left(\dfrac{3}{5}\right)^2\right]$

i.e. $\;$ $b^2 = 25 \times \left[\dfrac{16}{25}\right]$ $\implies$ $b^2 = 16$

$\therefore$ $\;$ Required equation of ellipse is

$\dfrac{x^2}{5^2} + \dfrac{y^2}{16} = 1$

i.e. $\;$ $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

Analytical Geometry - Conics - Ellipse

Find the equation of the ellipse if one of the foci is $\left(0,-1\right)$, the corresponding directrix is $\;$ $3x + 16 = 0$ $\;$ and $\;$ $e = \dfrac{3}{5}$

Let $\;$ $P \left(x,y\right)$ $\;$ be a moving point.

Focus $= F = \left(0,-1\right)$

Equation of directrix is $\;$ $3x + 16 = 0$

Draw $\;$ $PM$ $\;$ perpendicular to the directrix.

By definition, $\;$ $\dfrac{FP}{PM}= e$

$\therefore$ $\;$ $FP^2 = e^2 \; PM^2$

i.e. $\;$ $\left(x - 0\right)^2 + \left(y + 1\right)^2 = \left(\dfrac{3}{5}\right)^2 \times \left(\dfrac{\left|3x + 16\right|}{\sqrt{3^2}}\right)^2$

i.e. $\;$ $x^2 + y^2 + 2y + 1 = \dfrac{9}{25} \times \left(\dfrac{9x^2 + 96x + 256}{9}\right)$

i.e. $\;$ $25x^2 + 25y^2 + 50y + 25 = 9x^2 + 96x + 256$

i.e. $\;$ $16x^2 + 25y^2 - 96x + 50y - 231 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the equation of the required ellipse.