Question

The tangent to the parabola $y=x^{2}-2x+8$ at $P\left(2 , 8\right)$ touches the circle $x^{2}+y^{2}+18x+14y+\lambda =0$ at $Q.$ The coordinates of point $Q$ are

• A. $\left(- 7 , - 12\right)$
• B. $\left(- 9 , - 13\right)$
• C. $\left(- 11 , - 16\right)$
• D. $\left(- \frac{31}{5} , - \frac{42}{5}\right)$

Answer 1

Answer: (D)

The equation of the tangent at $P \, \left(2 , 8\right)$ to the parabola is
$\frac{y + 8}{2}=2x-\left(x + 2\right)+8$
$\Rightarrow 2x-y+4=0 \, \, ...\left(\right.i\left.\right)$
So, the slope of the normal to the circle is $-\frac{1}{2}$
The equation of normal to the circle at $Q$ will be
$y+7=-\frac{1}{2}\left(x + 9\right)$
$x+2y+23=0 \, \, ...\left(\right.ii\left.\right)$
The point $Q$ is the intersection of the tangent at $P$ and the normal at $Q$
On solving the equation $\left(\right.i\left.\right) \, \& \, \left(\right.ii\left.\right)$ , we get,
$x+2\left(2 x + 4\right)+23=0$
$\Rightarrow 5x=-31\Rightarrow x=-\frac{31}{5}$
and $y=2\left(- \frac{31}{5}\right)+4=-\frac{42}{5}$
Hence, the coordinates of the point $Q$ are $\left(- \frac{31}{5} , - \frac{42}{5}\right)$