Question

If CF is the perpendicular from the centre C of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ on the tangent at any point P, and G is the point when the normal at P meets the major axis, then CF. PG =

• A. $a^{2}$
• B. $ab$
• C. $b^{2}$
• D. $b^{3}$

Answer 1

Answer: (C)

$b^{2}$

Answer 2

Ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Tangent at 'P' is $\frac{x}{a}\cos\varphi + \frac{y}{b}\sin\varphi = 1$

$\therefore CF = \frac{1}{\sqrt{\left( \frac{\cos^{2}\varphi}{a^{2}} + \frac{\sin^{2}\varphi}{b^{2}} \right)}}$

=$\frac{ab}{\sqrt{\left( a^{2}\sin^{2}\varphi \right) + b^{2}\cos^{2}\varphi}}$

Equation of normal at P

=$ax\sec\varphi - by\cos ec\varphi = a^{2} - b^{2}$then $G = \left( \frac{\left( a^{2} - b^{2} \right)}{a}\cos\varphi,0 \right)$

$\therefore PG = \sqrt{\left( a\cos\varphi - \frac{\left( a^{2} - b^{2} \right)}{a}\cos\varphi \right)^{2} + b\left( \sin\varphi - 0 \right)^{2}}$=

$\sqrt{\left( \frac{b^{4}}{a^{2}}\cos^{2}\varphi + b^{2}\sin^{2}\varphi \right)}$

=$\frac{b}{a}\sqrt{\left( a^{2}\sin\varphi + b^{2}\cos^{2}\varphi \right)}$

⇒ $CF.PG = b^{2}$