# Find the equation of the ellipse with eccentricity

Question:

Find the equation of the ellipse with eccentricity $\frac{3}{4}$, foci on the $y$-axis, center at the origin and passing through the point (6, 4).

Solution:

Given that

Eccentricity $=\frac{3}{4}$

we know that,

Eccentricity, $\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}$

$\Rightarrow \frac{3}{4}=\frac{c}{a}$

$\Rightarrow c=\frac{3}{4} a$

We know that,

$c^{2}=a^{2}-b^{2}$

$\Rightarrow\left(\frac{3 a}{4}\right)^{2}=a^{2}-b^{2}$

$\Rightarrow \frac{9 a^{2}}{16}=a^{2}-b^{2}$

$\Rightarrow b^{2}=a^{2}-\frac{9 a^{2}}{16}$

$\Rightarrow b^{2}=\frac{16 a^{2}-9 a^{2}}{16}$

$\Rightarrow \mathrm{b}^{2}=\frac{7 \mathrm{a}^{2}}{16}$ ................(i)

It is also given that Coordinates of foci is on the $y$-axis

So, Equation of ellipse is of the form

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

Substituting the value of $b^{2}$ in above eq., we get

$\frac{x^{2}}{\frac{7 a^{2}}{16}}+\frac{y^{2}}{a^{2}}=1$

$\Rightarrow \frac{16 x^{2}}{7 a^{2}}+\frac{y^{2}}{a^{2}}=1$ …(ii)

Given that ellipse passing through the points (6, 4)

So, point $(6,4)$ will satisfy the eq. (ii)

Taking point $(6,4)$ where $x=6$ and $y=4$

Putting the values in eq. (ii), we get

$\frac{16(6)^{2}}{7 a^{2}}+\frac{(4)^{2}}{a^{2}}=1$

$\Rightarrow \frac{16 \times 36}{7 \mathrm{a}^{2}}+\frac{16}{\mathrm{a}^{2}}=1$

$\Rightarrow \frac{576+7 \times 16}{7 \mathrm{a}^{2}}=1$

$\Rightarrow \frac{576+112}{7 \mathrm{a}^{2}}=1$

$\Rightarrow \frac{688}{7 \mathrm{a}^{2}}=1$

$\Rightarrow \mathrm{a}^{2}=\frac{688}{7}$

Substituting the value of $a^{2}$ in eq. (i), we get

$b^{2}=\frac{7 \times \frac{688}{7}}{16}$

$\Rightarrow b^{2}=\frac{688}{16}$

Substituting the value of $a^{2}$ and $b^{2}$ in the equation of an ellipse, we get

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

$\Rightarrow \frac{x^{2}}{\frac{688}{16}}+\frac{y^{2}}{\frac{688}{7}}=1$

$\Rightarrow \frac{16 x^{2}}{688}+\frac{7 y^{2}}{688}=1$

or $16 x^{2}+7 y^{2}=688$