Question

If the distances of two points $P$ and $Q$ which lie on a parabola $y^2=4ax$ from focus are $3$ and $12$ units respectively then distance of the point of intersection of tangents at $P$ and $Q$ from the focus is

Answer 1

If distance of two points P and Q which lie on a parabola $y^2=4ax$ from focus are $3$ and $12$ units then distance of point of intersection of tangent at P and Q from focus is :

Distance $(a{t}^2+a{)}^2+(2at{)}^2=9$

$(a{t}^2+a{)}^2=9t^2=\frac{3}{a}-1t_1=\pm \sqrt{\frac{3}{a}-1}$

$t_2=\pm \sqrt{\frac{12}{a}-1}$

$P=(a{t}_1^2,2a{t}_1)=\left[\right(3-\left)2\sqrt{a(3-a)}\right]=Q=\left[\right(12-a),2\sqrt{a(12-a)}]$

Tangent at P and Q are

$2\sqrt{a(3-a)y}=2ax+2a(3-a).........\left(1\right)2\sqrt{a(12-a)}=2ax+2a(12-a).........\left(2\right)$

Solve (1) and (2)

$2\sqrt{a(3-a)y}=2ax+2a(3-a).........\left(1\right)2\sqrt{a(12-a)}=2ax+2a(12-a).........\left(2\right)x=\sqrt{(3-a)(12-a)}y=\sqrt{a}[\sqrt{12-a}+\sqrt{3-a}]$

Let intersection point is $(x,y)$ and focus $(a,0)$

Distance between $(x,y)$ and $(a,0)$ is

$\sqrt{({(\sqrt{(3-a)(12-a)}-a)}^2+{(\sqrt{a}(\sqrt{12-a}+\sqrt{3-a})-0)}^2)}=36$

$=6$ Units.