Question

Normal to the parabola $y^2=8x$ at the point $p(2,4)$ meets the parabola again at the point $Q.$ If $C$ is the center of the circle described on $PQ$ as diameter then the coordinates of the image of the point $C$ in the line $y=x$ are

• A. $(-4,10)$
• B. $(-3,8)$
• C. $(4,-10)$
• D. $(-3,10)$

Answer 1

The correct option is A $(-4,10)$

Equation of the normal at point $(x_1,y_1)$ on parabola $y^2=4ax$ is given by

$y-y_1=\frac{-y_1}{2\times a}(x-x_1)$

The equation of the normal at point (2,4)

$y-4=\frac{-4}{2\times 2}(x-2)$.

$⟹y-4=2-x⟹x=6-y$

To find the other point of the intersection of the parabola and normal from the equation of the parabola $y^2=8(6-y)⟹y^2+8y-48=0y=-12,4$

The x coordinate of point is $8x={12}^2=144⟹x=18$

therefore the endpoints of the diagonal of the circle are $(2,4)$ and $(18,-12)$

Centre of the circle is $C(\frac{18+2}{2},\frac{4-12}{2})=(10,-4)$

Image of the C in line $x=y$ is $(-4,10)$