JEE-Advanced 2015 (2/40)

Parametric point on this parabola is : $(t^2 , 2t)$ . Now by the formula of finding the point of reflection of point $(x_1,y_1)$ about the line $L : ax+by+c=0$

$\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{-2(ax_1 + by_1 +c)}{a^2+b^2}$

we get

$\dfrac{x-t^2}{1} = \dfrac{y-2t}{1} = \dfrac{-2(t^2+2t+4)}{2} \implies x = -2t - 4 \ , \ y = -t^2 - 4$

This is the parametric form of the curve $C$ given by

$C : x^2 + 4y + 8x + 32 = 0$

(the equation can be derived by isolating $t$ and $t^2$ in the parametric equations and performing algebraic manipulations to eliminate $t$ .)

Now, the point of intersection of the curve with $y=-5$ can be found out by putting $y=-5$ in $C$ to obtain

$x^2 - 20 + 8x + 32 = 0 \implies x^2 + 8x + 12 = 0 \implies (x+6)(x+2) = 0 \implies x = -6, -2$

Thus the required points of intersection are : $(-6,-5)$ and $(-2,-5)$ and the distance between these two points is $\boxed{4}$ units.

Find the image of line y=-5 in the line x-y=8 and find its intersection points with C. Required distance is same as the distance between these points.