Question

Find the equation of an ellipse with its foci on y-axis, eccentricity 3/4, centre at the origin and passing through (6,4).

Answer 1

Let the equation of the required ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where a Now,
$a^2=b^2(1-e^2)$
$\Rightarrow a^2=b^2(1-e^2$
$\Rightarrow a^2=b^2[1-{\left(\frac{3}{4}\right)}^2]$
$\Rightarrow a^2=b^2[1-\frac{9}{16}]$
$\Rightarrow a^2=b^2\times \frac{7}{16}$
$\Rightarrow a^2=\frac{7}{16}{b}^2$ ...(ii)
The required ellipse through (6,4).
$\therefore \frac{(6{)}^2}{a^2}+\frac{(4{)}^2}{b^2}=1$
$\Rightarrow \frac{36}{a^2}+\frac{16}{b^2}=1$
$\Rightarrow \frac{{36}^2}{\frac{7}{16}{b}^2}+\frac{16}{b^2}=1$
$\Rightarrow \frac{36\times 36}{7b^2}+\frac{16}{b^2}=1$
$\Rightarrow \frac{576}{7b^2}+\frac{16}{b^2}=1$
$\Rightarrow \frac{1}{b^2}[\frac{576}{7}+\frac{16}{1}]=1$
$\Rightarrow \frac{576}{7}+\frac{16}{1}=b^2$
$\Rightarrow b^2=\frac{688}{7}$
Putting $b^2=\frac{688}{7}$ in equation (i), we get
$a^2=\frac{7}{16}\times \frac{688}{7}$
$\Rightarrow a^2=\frac{688}{16}=43$
Putting $a^2=43$ and $b^2=\frac{688}{7}$ in equation (i), we get,
$\frac{x^2}{43}+\frac{y^2}{\frac{688}{7}}=1$
$\Rightarrow \frac{x^2}{43}+\frac{7y^2}{688}=1$
This is the equation of the required ellipse.