Question

If $y=2x+1$ is axis of a parabola. Let $x+2y+3=0$ and $y=x+1$ are the two tangents of the parabola. If lengths of the latus rectum of the parabola is $\frac{\sqrt{p}}{q}$, where $p$ and $q$ are coprime then the value of $p+q$ is

Answer 1

Here, axis $y=2x+1$ and the tangent $x+2y+3=0$ are perpendicular to each other.
$\Rightarrow$ vertex of parabola is the intersection of $y=2x+1andx+2y+3=0$
$\Rightarrow$ Vertex : (-1,-1)
Also Focus lies on axis of parabola $y=2x+1$
$\Rightarrow$ Let $S(t,2t+1)$ be the focus
Property: The foot of perpendicular from the focus upon any tangent lies on the tangent at the vertex.

Let $(h,k)$ be the foot of the perpendicular from $S(t,2t+1)$ on the tangent $y=x+1$
$\frac{h-t}{1}=\frac{k-(2t+1)}{-1}=-\left(\frac{t-2t-1+1}{2}\right)$
$\Rightarrow h=t+\frac{t}{2}=\frac{3t}{2}$ &
$k=2t+1-\frac{t}{2}=\frac{3t}{2}+1$
Now $(h,k)$ lies on $x+2y+3=0$
$\Rightarrow h+2k+3=0$
$\Rightarrow \frac{3t}{2}+3t+2+3=0$
$\Rightarrow 3t+6t+10=0$
$\Rightarrow t=\frac{-10}{9}$
$S(\frac{-10}{9},\frac{-11}{9})$
As the distance between vertex and focus is the length a
$\Rightarrow$
$a=\sqrt{{(-1+\frac{10}{9})}^2+{(-1+\frac{11}{9})}^2}$
$\sqrt{\frac{1}{81}+\frac{4}{81}}=\frac{\sqrt{5}}{9}$
Length of latus rectum = $4a=4\frac{\sqrt{5}}{9}=\frac{\sqrt{80}}{9}=\frac{\sqrt{p}}{q}$
$p+q=89$