Question

If the latusrectum of an ellipse with axis along x-axis and centre at origin is 10, distance between foci = length of minor axis, then equation of the ellipse is ___________.

Answer 1

Centre of ellipse is given (0, 0)
Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Latusrectum is given, which is 10
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{2b^2}{a}=10 \\ \mathrm{i}.\mathrm{e}\frac{b^2}{a}=5...\left(1\right) \end{gathered}$$
Also given, distance between foci = Length of minor axis
i.e 2ae = 2b
i.e ae = b
i.e e = $\frac{b}{a}$
Since b² = a² (1 – e²)
i.e a²e² = a² (1 – e²)
i.e e² = 1 – e²
i.e 2e² = 1
i.e e = $\pm \frac{1}{\sqrt{2}}$
i.e e = $\frac{1}{\sqrt{2}}$
$\Rightarrow a=b\sqrt{2}$
Since $\frac{2b^2}{a}=10$
$$\begin{gathered} \Rightarrow \frac{2b^2}{b\sqrt{2}}=10 \\ \mathrm{i}.\mathrm{e}b=\frac{10}{\sqrt{2}} \\ \mathrm{i}.\mathrm{e}b=5\sqrt{2} \\ \Rightarrow a=10 \end{gathered}$$
$\therefore$ equation of ellipse is $\frac{x^2}{{\left(10\right)}^2}+\frac{b^2}{{\left(5\sqrt{2}\right)}^2}=1$
i.e $\frac{x^2}{100}+\frac{b^2}{50}=1$
i.e x² + 2b² – 100
Hence equation of ellipse in x² + 2b² = 100