Question

The normal at a variable point $P$ on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ meets axes of the ellipse at $Q$ and $R.$ Then which of the following is/are CORRECT?

• A. The locus of mid-point of $QR$ is an ellipse
• B. The locus of mid-point of $QR$ is a conic whose eccentricity is same as of given ellipse
• C. The locus of mid-point of $QR$ is a hyperbola
• D. The locus of mid-point of $QR$ is a conic with its eccentricity half of the eccentricity of given ellipse

Answer 1

Answer: (A), (B)

The locus of mid-point of $QR$ is a conic whose eccentricity is same as of given ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $(a>b)$
Equation of normal at point $P(a\cos \theta ,b\sin \theta )$ is,
$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$
It meets axes at $Q(\frac{(a^2-b^2)\cos \theta }{a},0)$ and $R(0,-\frac{(a^2-b^2)\sin \theta }{b})$

Let $M(h,k)$ be the midpoint of $QR.$
Then $2h=\frac{(a^2-b^2)\cos \theta }{a}$ and $2k=-\frac{(a^2-b^2)\sin \theta }{b}$
$\Rightarrow {\cos }^2\theta +{\sin }^2\theta =\frac{4h^2{a}^2}{(a^2-b^2{)}^2}+\frac{4k^2{b}^2}{(a^2-b^2{)}^2}=1$
Locus is $\frac{x^2}{{\left(\frac{a^2-b^2}{2a}\right)}^2}+\frac{y^2}{{\left(\frac{a^2-b^2}{2b}\right)}^2}=1$ which represents an ellipse.

Since $a^2{e}^2=a^2-b^2,$
$\therefore \frac{x^2}{\left(\frac{a^4{e}^4}{4a^2}\right)}+\frac{y^2}{\left(\frac{a^4{e}^4}{4b^2}\right)}=1$

Since $a>b,$
$\therefore \frac{a^4{e}^4}{4b^2}>\frac{a^4{e}^4}{4a^2}{e}^{\prime }=\sqrt{1-\frac{\left(\frac{a^4{e}^4}{4a^2}\right)}{\left(\frac{a^4{e}^4}{4b^2}\right)}}=\sqrt{1-\frac{b^2}{a^2}}=e\Rightarrow e^{\prime }=e$