Question

If the normal to a parabola $y^2=4ax$ at $P$ meets the curve again at $Q$ and if $PQ$ and the normal at $Q$ makes angle $\alpha$ and $\beta$, respectively with the $x$-axis then $\tan \alpha (\tan \alpha +\tan \beta )$ has the value equal to

• A. $0$
• B. $-2$
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

The correct option is B $-2$

$y^2=4ax$


Point on the parabola $(at,2at)$
Slope of the tangent,
$2y{y}^{\prime }=4a\Rightarrow y^{\prime }=\frac{1}{t}$
Slope of the normal,
$m_n=-t$
So,
$\tan \alpha =-t_1$ and $\tan \beta =-t_2$
Also, $t_2=-t_1-\frac{2}{t_1}$
$t_1{t}_2+t_1^2=-2$
$\tan \alpha \tan \beta +{\tan }^2\alpha =-2$