Question

Find the equation of an ellipse whose axes lie along coordinate axes and which passes through (4,3) and (-1,4).

Answer 1

Let the equation of the required ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where a>b ...(i)
The required ellipse passes through (4,3) and (-1,4)
$\therefore \frac{(4{)}^2}{a^2}+\frac{(3{)}^2}{b^2}=1$
$\frac{16}{a^2}+\frac{9}{b^2}=1$
$\Rightarrow 16b^2+9a^2=a^2{b}^2$ ...(ii)
and $\frac{(-1{)}^2}{a^2}+\frac{(4{)}^2}{b^2}=1$
$\Rightarrow \frac{1}{a^2}\times \frac{16}{b^2}=1$
$\Rightarrow b^2+16a^2=a^2{b}^2$ ...(iii)
Multiplying equation (iii) by 16, we get
$16b^2+256a^2=16a^2{b}^2$ ...(iv)
Substracting equation (ii) from equation (iv), we get
$256a^2-9a^2=16a^2{b}^2-a^2{b}^2$
$\Rightarrow 247a^2=15a^2{b}^2$
$\Rightarrow \frac{247}{15}=b^2$
$\Rightarrow b^2=\frac{247}{15}$
Putting $b^2=\frac{247}{15}$ in equation (iii), we get
$\frac{247}{15}+16a^2=a^2\times \frac{247}{15}$
$\Rightarrow 16a^2-\frac{247a^2}{15}=\frac{-247}{15}$
$\Rightarrow \frac{240a^2-247a^2}{15}=\frac{-247}{15}$
$\Rightarrow -7a^2=-247$
$\Rightarrow a^2=\frac{247}{7}$
Putting $a^2=\frac{247}{7}$ and $b^2=\frac{247}{15}$ in equation (i), we get,
$\frac{x^2}{\frac{247}{7}}+\frac{y^2}{\frac{247}{15}}=1$
$\Rightarrow \frac{7x^2}{247}+\frac{15y^2}{247}=1$
This is the equation of the required ellipse.