Question

find the equation of ellipse whose vertices are (0, +-10) and eccentricity 4/5

Answer 1

Answer:

$\frac{ {x}^{2} }{36} + \frac{ {y}^{2} }{100} = 1 \\ \:$

Step-by-step explanation:

We know that standard equation of ellipse is

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

here vertices A(0,-a) and B(0,a)

To find the equation of ellipse whose vertices are (0, +-10) and eccentricity 4/5 ,

here

$a = 10 \\ e = \frac{c}{a} = \frac{4}{5} \\ \\ c = ae \\ \\ c = 10 \times \frac{4}{5} = 8$

now

${c}^{2} = {a}^{2} - {b}^{2} \\ \\ {b}^{2} = {a}^{2} - {c}^{2} = 100 - 64 = 36 \\ \\ {b}^{2} = 36 \\ \\ {a}^{2} = 100$

So,equation if ellipse

$\frac{ {x}^{2} }{36} + \frac{ {y}^{2} }{100} = 1$

Hope it helps you.

Answer 2

Correct Question :-

Find the equation of ellipse for which e=4/5 and whose vertices are (0,±10)

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Here we have,

Vertices of ellipse lie on the y axis (Given)

Hence,

It is a Vertical ellipse

Now,

Assume

${\boxed{\sf\:{Equation=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}}}$

Where a² > b²

Vertices = (0 , ±a)

Hence,

a = 10

Assume

c² = (a² - b²)

Then,

$\tt{\rightarrow e=\dfrac{c}{a}}$

c = ae

$\tt{\rightarrow 10\times\dfrac{4}{5}}$

= 8

Now,

c² = (a² - b²)

b² = (a² - c²)

= 100 - 64

Also

= a²

= (10)²

= 100

b² = 36

Hence

$\Large{\boxed{\sf\:{Equation=\dfrac{x^2}{36}+\dfrac{y^2}{100}=1}}}$