Question

$PQ$ is a double ordinate of the ellipse $x^2+9y^2=9$, the normal at $P$ meets the diameter through $Q$ at $R$, then the locus of the midpoint of $PR$ is -

• A. a circle
• B. a parabola
• C. an ellipse
• D. a hyperbola

Answer 1

The correct option is C an ellipse

REF.Image.

A $\frac{x^2}{9}+\frac{y^2}{1}=1$

$P(3cos\theta ,sin\theta )$

Equation of normal at 'P' is

$3xsec\theta -ycosec\theta =8$ __ (1)

$\Rightarrow 3xsec\theta -ycosec\theta =8$ __ (1)

Equation of Diameter

$\frac{y-0}{x-0}=\frac{sin\theta }{3cos\theta }$

$\Rightarrow y=\frac{-xsin\theta }{3cos\theta }$ __(2)

$3xsec\theta +x\frac{sin\theta }{3cos\theta }cosec\theta =8$

$9x+x=24cos\theta$

$x=\frac{24}{10}cos\theta$ __ (3)

$\therefore y=\frac{-8}{10}sin\theta$ __ (4)

$R(\frac{24cos\theta }{10},\frac{-8sin\theta }{10}),P(3cos\theta ,sin\theta )$

$\therefore h=\frac{3cos\theta +\frac{24}{10}cos\theta }{2},$ $k=\frac{\frac{-8}{10}sin\theta +sin\theta }{2}$

$\therefore cos\theta =\frac{10h}{27},$ $10k=sin\theta$

$\therefore \frac{h^2}{(\frac{24}{10}{)}^2}+\frac{k^2}{(\frac{1}{10}{)}^2}=1$

Hence, locus is ellipse.