Question

The condition that the two tangents to the parabola $y^2=4ax$ become normal to the circle $x^2+y^2-2ax-2by+c=0$ is given by

• A. $a^2>4b^2$
• B. $b^2>2a^2$
• C. $a^2>2b^2$
• D. $b^2>4a^2$

Answer 1

The correct option is D $b^2>4a^2$

we have

$x^2+y^2-2ax-2by+c=0$

center will be $\{-\frac{(-2a)}{2},-\frac{(-2b)}{2}\}=(a,b)$

Pain of tangents will par through center of circle or

in other wads, per of length must originate from

center of circle

we have

$y^2=4acx$

$S:y^2-4ax=0$

$(x_1,y_1)=(a,b)$

$S_1{y}_1^2-4a{x}_1=0$

$T:y{y}_1-4a\left(\frac{x_1+x}{2}\right)=0$

$\Rightarrow T=y{y}_1-2a(x-x_1)=0$

Equation of pan of tangents is :

$S{S}_1=T_2$

$\Rightarrow (y^2-4ax)(y_1^2-4a{x}_1)={\{y{y}_1-2a(x+x_1\left)\right\}}^2$

$\Rightarrow (y^2-4ax)(b^2-4a.a)={\{yb-2a(x+a\left)\right\}}^2$ on substituent $(x_1,y_2)(a,b)$

$\Rightarrow (y^2-4ax)(b^2-4a^2)={\{by-2a(x+a\left)\right\}}^2$

from here, $b^2$ should be grater then $4a^2$

then $b^2>4a^2$