Question

If CF is perpendicular form the center of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to the tangent at P, and G is the point where the normal at P meets the major axis, then the product, CF . PG is

• A. $a^2$
• B. $2b^2$
• C. $b^2$
• D. $a^2-b^2$

Answer 1

Answer: (C)

$b^2$

We have,

CF is perpendicular Form the center

the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

to tangent at the price P.

Let $P(acos\theta .bsin\theta )$ on the ellipse is

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

then, we know that,

the equation of tangent is

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

Now,

$\frac{xacos\theta }{a^2}+\frac{ybsin\theta }{b^2}=1$

$\Rightarrow \frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$

$\Rightarrow bxcos\theta +aysin\theta =ab$

but CF be the perpendicular

then

$CF=\left|\frac{ab}{\sqrt{b^{2}co{s}^2\theta +a^{2}si{n}^2\theta }}\right|...\left(1\right)$

Equation of Normal at point

$P(acos\theta ,bsin\theta )$ on the ellipse is

$\frac{ax}{cos\theta }-\frac{by}{sin\theta }=a^2-b^2$

Solving with $y=0$, we get

$G=(\frac{(a^2-b^2)cos\theta }{a},0)$

using distance formula,

$PG=\frac{b}{a}\sqrt{b^{2}co{s}^2\theta +a^{2}si{n}^2\theta }...\left(2\right)$

on multiplying by equation (1) and (2) to

and we get,

$CF.PG=b^2$