Question

Let $PQ$ be a focal chord of $y^2=4ax$. The tangents to parabola at $P$ and $Q$ meet at a point lying on the line $y=2x+a(a>0)$. If the chord $PQ$ subtends an angle $\theta$ at the vertex of prabola then $\tan \theta =$

• A. $\frac{2\sqrt{7}}{3}$
• B. $\frac{2\sqrt{7}}{5}$
• C. $\frac{2\sqrt{5}}{3}$
• D. $\frac{2\sqrt{3}}{5}$

Answer 1

The correct option is C $\frac{2\sqrt{5}}{3}$

Let $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$
For focal chord $t_1{t}_2=-1$
Let $O$ be the vertex, then slope of $PO=m_1=\frac{2}{t_1}$
And slope of $QO=m_2=\frac{2}{t_2}$
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{2(t_2-t_1)}{t_1{t}_2+4}\right|$
Let point of intersection of tangents at $t_1,t_2$ be $R=(a{t}_1{t}_2,a(t_1+t_2\left)\right)$
$R$ lies on the directix $x=-a$ and it also lies on line $y=2x+a$
$\therefore y=-a=a(t_1+t_2)$
$\Rightarrow t_1+t_2=-1$
$(t_2-t_1{)}^2=(t_1+t_2{)}^2-4t_1{t}_2=5$
$t_2-t_1=\pm \sqrt{5}$
$\tan \theta =\left|\frac{2\sqrt{5}}{-1+4}\right|=\frac{2\sqrt{5}}{3}$