Question

The locus of mid points of the chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose poles lie on the auxiliary circle is:

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

Answer 1

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

Let $P(x_1,y_1)$ be a point in a locus. Equation of the chord having $P$ as its midpoint is $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}---\left(i\right)$

Let $Q$ be the pole. $Q$ lies on the auxiliary circle i.e, $x^2+y^2=a^2$

So, $Q=(a\cos \theta ,a\sin \theta )$

The polar of $Q$ with respect to the ellipse is $\frac{x\left(a\cos \theta \right)}{a^2}+\frac{y\left(a\sin \theta \right)}{b^2}=1---\left(ii\right)$

comparing coefficients in $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow \frac{\frac{a\cos \theta }{a^2}}{\frac{x_1}{a^2}}=\frac{\frac{a\sin \theta }{b^2}}{\frac{y_1}{b^2}}=\frac{1}{[\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}]}$

$\Rightarrow \cos \theta =\frac{\frac{x_1}{a}}{[\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}]},\sin \theta =\frac{\frac{y_1}{a}}{[\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}]}{\cos }^2\theta +{\sin }^2\theta =1\Rightarrow \frac{x_1^2}{a^2}+\frac{y_1^2}{a^2}=(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}{)}^2$

the locus of midpoints is, $\frac{x^2+y^2}{a^2}=(\frac{x^2}{a^2}+\frac{y^2}{b^2}{)}^2$