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...

Question

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?

Answer 1

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.

Answer 2

As you rightly obtained, $t^2 = 2 \implies t = \pm \sqrt2$

If we take $t = \sqrt2$ and rewrite,

$P(2a, 2a \sqrt2), Q(\frac{a}{2}, - a \sqrt2)$

Slope of tangent at point $Q = \frac{2a}{y} = - \sqrt2$

So equation of tangent line $y + a \sqrt2 = -\sqrt2(x-\frac{a}{2})$

or $y = - \sqrt2 x - \frac{a}{\sqrt2} $

Point $R$ where the line intersects the directrix, $R(-a, \frac{a}{\sqrt2})$.

So $QR^2 = \frac{9a^2}{4} + \frac{9a^2}{2} = 36 \implies a = \frac{4}{\sqrt3}$.

Now slope of tangent line at $P = \frac{2a}{y} = \frac{1}{\sqrt2}$.

So equation of tangent line at $P$, $y - 2a\sqrt2 = \frac{1}{\sqrt2}(x - 2a)$

or $y = \frac{x}{\sqrt2} + \sqrt2 a$

Its distance from focus $(a,0)$ is

$d = \frac{3a/ \sqrt2}{\sqrt{3/2}} = a \sqrt3 = 4$

Answer 3

You don't need to use coordinates for this problem. Line $PR$ is the tangent at $P$ and triangle $PQR$ is right-angled at $R$. Distance $FH$ is thus parallel to $QR$. By similar triangles: $$ FH:QR=FP:QP=2:3, \quad\text{hence:}\quad FH={2\over3}QR=4. $$