Question

Find the equation of ellipse with eccentricity 3/4 center at origin foci on y axis passing through point (6,4)

Answer 1

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

Here we have,

Foci of ellipse lie on the y axis (Given)

Hence,

It is a Vertical ellipse

Now,

Assume

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

Where a² > b²

Assume

c² = (a² - b²)

$\tt{\rightarrow e=\dfrac{3}{4}}$

$\tt{\rightarrow\dfrac{c}{a}=\dfrac{3}{4}}$

$\tt{\rightarrow c=\dfrac{3}{4}a}$

Therefore,

b² = (a² - c²)

$\tt{\rightarrow a^2-\dfrac{9}{16}a^2}$

$\tt{\rightarrow\dfrac{7a^2}{16}}$

Hence,

$\Large{\boxed{\sf\:{Equation=\dfrac{x^2}{7a^2/16}+\dfrac{y^2}{a^2}=1}}}$

16x² + 7y² = 7a² ..... (2)

It passes through (6,4) put value in (2)

7a² = 688

Therefore

Required Equation = 16x² + 7y² = 688