Let the focus S of parabola divide one of its focal chord in the ratio 2:1. If the tangent at Q cuts directrix at R such that RQ=6, find distance of focus from tangent at P.
Let P and Q be $(at^2,2at)$ and $(\frac{a}{t^2}, \frac{-2a}{t})$ respectively, for parabola $y^2=4ax$
So $$0= 2(-\frac{2a}{t})+2at$$
$$\implies t^2=2$$
Now tangent at Q is
$$\frac{y}{\sqrt 2} =x+\frac a2$$
So point R is $(-a, -\frac{a}{\sqrt 2})$
Then $$QR^2 = 36$$
$$\implies a=\frac{12}{\sqrt {11}}$$
So now, tangent at P is $$\sqrt 2y=x+2a$$ whose distance from $(a,0)$ is
$$d=\left | \frac{a+2a}{\sqrt 3}\right|$$
$$d=\sqrt 3 a$$
$$d=\frac{\sqrt 3 \times 12}{\sqrt {11}}$$
But given answer is 4
Where am I going wrong?
Best Answer
The mistake lies in taking incorrect sign of $t$ for point $Q$. It lies on opposite side of $P$. So for $Q$, parameter is $-1/t=-1/\sqrt{2}$ but in the tangent equation at $Q$ you took $1/t=1/\sqrt{2}$ which gave wrong coordinates for $R$.
Since $(x,y)=(at^2,2at)$, $t$ has same sign as $y$. So the entire upper branch of a parabola has $t$ positive and entire lower branch has $t$ negative - something to keep in mind.