Question

The locus of the mid-points of chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ that touch the circle $x^2+y^2=b^2$, is:

• A. ${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2=\frac{x^2}{a^4}+\frac{y^2}{b^4}$
• B. ${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2=b^2(\frac{x^2}{a^4}+\frac{y^2}{b^4})$
• C. ${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2=a^2(\frac{x^2}{a^4}+\frac{y^2}{b^4})$
• D. None of these

Answer 1

The correct option is B ${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2=b^2(\frac{x^2}{a^4}+\frac{y^2}{b^4})$

equation of chord with given mid point $(x_1,y_1)$ is given by $T=S_1$
$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=\frac{x_1^2}{a^2}+\frac{{y_1}^2}{b^2}$
Since, it touches the circle $x^2+y^2=r^2$

So, its' distance from the centre of circle is $r$
therefore, $r=p$

$\Rightarrow r=\left|\frac{\frac{x_1^2}{a^2}+\frac{{y_1}^2}{b^2}}{\sqrt{\frac{x_1^2}{a^4}+\frac{{y_1}^2}{b^4}}}\right|$
Therefore, locus is ${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2=r^2(\frac{x^2}{a^4}+\frac{y^2}{b^4})$

Put $r=b,$ we get option $B$ as answer.