Question

Let $A$ and $B$ be two distinct points on the parabola $y^2=4x$. If the axis of the parabola touches a circle of radius $2$ having $AB$ as its diameter, Then the slope of the line joining $A$ and $B$ can be

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. $Noneofthese$

Answer 1

Answer: (C)

$1$

General point on the parabola $y^2=4x$ is $(t^2,2t)...$

parabola cuts the circle at two different points, let these points be $A$ & $B$.

Say, the coordinates of $A(t_1^2,{2t}_1)$ & $B(t_2^2,{2t}_2)$

also given n that the circle touches the axis of the parabola i.e. touches the $x-$axis, so the $y-$coordinate of the center is $+r$ when the circle is drawn above the $X-$axis & ${}^{\prime }-r^{\prime }$ when drawn below.

And the radius is given to be $r=2$ units

$\because$ $AB$ is the diameter then

Coordinate of center of circle $=\frac{A+B}{2}=[\frac{(t_1^2+t_2^2)}{2},\frac{(t_1+t_2)2}{2}]=(\frac{t_1^2+t_2^2}{2},t_1+t_2)$

Now the $y-$coordinate of the center $=\pm 2$

$\therefore$ $t_1+t_2=2$ or $t_1+t_2=-2$

Now slope of the line segment $AB$ is

$m=\frac{2(t_2-t_1)}{t_2^2-t_1^2}=\frac{2(t_2-t_1)}{(t_2-t_1)(t_2+t_1)}=\frac{2}{(t_2+t_1)}$

$\therefore$ $m=\frac{2}{2}$ or $\frac{2}{-2}=1$ or $-1$