Question

A tangent to the hypererbola $\frac{x^2}{4}-\frac{y^2}{1}=1$ meets ellipse $x^2+4y^2=4$ at two distinct points. The locus of mid point of this chord is

• A. ${\left(x^2+4y^2\right)}^2=4\left(x^2-4y^2\right)$
• B. $\left(x^2-4y^2\right)=4\left(x^2+4y^2\right)$
• C. ${\left(x^2-4y^2\right)}^2=4\left(4x^2+y^2\right)$
• D. ${\left(x^2+4y^2\right)}^2=4\left(4x^2-y^2\right)$

Answer 1

Answer: (A)

Any tangent to given hypererbola, is
$\frac{x\cdot 2\sec \theta }{4}-\frac{y\cdot \tan \theta }{1}=1$
Let $\left(x_1,y_1\right)$ be the middle point of the chord of ellipse.
$\therefore$ Its equation is $\frac{x{x}_1}{4}+\frac{y{y}_1}{1}=\frac{x_1^2}{4}+\frac{y_1^2}{1}$
As (1) and (2) are identical, so
$\frac{2\sec \theta }{x_1}=\frac{-\tan \theta }{y_1}=\frac{1}{\frac{x_1^2}{4}+\frac{y_1^2}{1}}\Rightarrow \sec \theta =\frac{2x_1}{x_1^2+4y_1^2}\text{ and }\tan \theta =\frac{-4y_1}{x_1^2+4y_1^2}$
As, ${\sec }^2\theta -{\tan }^2\theta =1$, so $\frac{4x_1^2}{{\left(x_1^2+4y_1^2\right)}^2}-\frac{16y_1^2}{{\left(x_1^2+4y_1^2\right)}^2}=1$
$\Rightarrow$ Locus is ${\left(x^2+4y^2\right)}^2=4\left(x^2-4y^2\right)$.