Question

find the equation of an ellipse that satisfies the given conditions vertices (0,±13) foci(0,±5).....pls answer ASAP

Answer 1

Hii dear here is your answer



Step 1 :

It is given that

Foci is (±3,0)(±3,0) and a = 4.

Since the foci lies on the x - axis, the major axis should be along x - axis.

Hence the equation of ellipse should be of the form

x2a2x2a2+y2b2+y2b2=1=1

c2=a2−b2c2=a2−b2 or b2=a2−c2b2=a2−c2

Now substituting the values for c and a we get,

b2=16−9b2=16−9

=7=7

Hence the equation of the ellipse is

x216x216+y27+y27=1






Hope it's help u

Answer 2

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

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 = 13

Assume,

Foci = (0 , ±c)

Hence,

c = 5

Also

b² = (a² - c²)

b² = (169 - 25)

= 144

b² = 144

Also

= a²

= (13)²

= 169

Hence

$\Large{\boxed{\sf\:{Equation=\dfrac{x^2}{144}+\dfrac{y^2}{169}=1}}}$