Question

Find the equation of the ellipse with vertices (0, ±13) ; foci (0, ±5)​

Answer 1

Vertices (0, +13), foci (0,+5)

Here, the vertices are on the y-axis.

Therefore, the equation of the ellipse will be of the form

$\frac{x {}^{2} }{b {}^{2} } + \frac{y {}^{2} }{ {a}^{2} } = 1$

, where a is the

semi-major axis.

Accordingly, a = 13 and c = 5.

It is known that a² =b²+c²

=> 169 = b²+25

=> b²=169-25

=> b²= 144

Thus, the equation of the ellipse is

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

Answer 2

Given:

$\red\bigstar\rm{Vertices =(0,\pm 13)}$ ......(1)

$\red\bigstar\rm{Foci=(0,\pm 5)}$ .........(2)

To Find :

• Equations of the ellipse =?

Solution :

Hence,

The vertices are of the form $\rm{(0,\pm a) }$

Hence, the major axis is along y axis

& Equation of ellipse is of the form

$\\ \blue \bigstar \boxed{ \sf{ \frac{ {x}^{2} }{ {b}^{2} } + \frac{ {y}^{2} }{ {a}^{2} } = 1}}$

From (1) & (2)

$\boxed{\bf{a=13}}$

Also, Given coordinate of foci $\rm{=(0,\pm 5)}$

We know that ,foci are $\rm{=(0,\pm c) }$

$\bf{So , \boxed{\bf{c=5}}}$

We know that,

$\: \implies \sf \: {c}^{2} = {a}^{2} - {b}^{2} \\ \\ \implies \sf \: {(5)}^{2} = {(13)}^{2} - {b}^{2} \\ \\ \implies \sf \: {b}^{2} = {(13)}^{2} - {(5)}^{2} \\ \\ \implies \sf \: {b}^{2} = 169 - 25 \\ \\ \implies \boxed{ \sf{{b}^{2} = 144 }}$

Equation of ellipse is :

$\\ \implies \sf \: \frac{ {x}^{2} }{ {b}^{2} } + \frac{ {y}^{2} }{ {a}^{2} } = 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \bf{ \bullet \: putting \: values \: }} \\ \\ \implies \sf \: \frac{ {x}^{2} }{144} + \frac{ {y}^{2} }{169} = 1 \\ \\ \\ \implies \boxed{ \sf{ \frac{ {x}^{2} }{144} + \frac{ {y}^{2} }{169} = 1 \: is \: the \: equation \: of \: ellipse}}$