Question

Let $E_1$ and $E_2$ be two ellipses $\frac{x^2}{a^2}+y^2=1$ and $x^2+\frac{y^2}{a^2}=1$ (where a is parameter) the locus of points of intersection of the ellipses $E_1$ and $E_2$ is a set of curves

• A. $y=x,y=-x,x^2+y^2=1$
• B. $y=2x,y=-2x,x^2+y^2=4$
• C. $(4x^2-y^2)(x^2+y^2-4)=0$
• D. $(x^2-y^2)(x^2+y^2-1)=0$

Answer 1

The correct option is D $(x^2-y^2)(x^2+y^2-1)=0$

$\begin{array}{c}Let\\ \frac{x^2}{a^2}+y^2=1-----\left(1\right)\\ x^2+\frac{y^2}{a^2}=1-----\left(2\right)\\ equation\left(1\right)-\frac{1}{a^2}.equ\left(2\right),weget\\ y^2-\frac{y^2}{a^4}=1-\frac{1}{a^2}\\ y^2\left(\frac{a^4-1}{a^4}\right)=\left(\frac{a^2-1}{a^2}\right)\\ y^2=\frac{a^2-1}{a^2}.\frac{a^4}{\left(a^2+1\right)\left(a^2-1\right)}\\ =\frac{a^2}{a^2+1}\\ and,\\ x=\pm \frac{a}{\sqrt{a^2+1}}\\ then,thelocusatpoint\\ \left(x^2-y^2\right)\left(x^2+y^2-1\right)=0\\ Hence,theoptionDisthecorrectanswer.\end{array}$