Question

A circle with its centre at the focus of the parabola $y^2=4ax$ and touching its directrix cuts the parabola at points $A$ & $B$, then length $AB$ is equal to

• A. $2a$
• B. $3a$
• C. $a$
• D. $4a$

Answer 1

The correct option is D $4a$

The circle with centre at focus of the parabola and touching its directrix will be,

${(x-a)}^2+{(y-0)}^2={\left(2a\right)}^2$
$x^2+a^2-2ax+y^2=4a^2$

For point of intersection, put $y^2=4ax$
$\therefore {(x+a)}^2=4a^2$
$(x+a)=2a$
$x=a(\because x>0)$
$y=-2a$ or $y=2a$

Therefore, the points are
$A(a,2a),B(a,-2a)$
$\therefore AB=4a$