Question

Suppose the parabola $(y–k{)}^2=4(x–h)$, with vertex $A$, passes through $O=(0,0)$ and $L=(0,2)$. Let $D$ be an end point of the latus rectum. Let the $y$-axis intersect the axis of the parabola at $P$. Then $\angle PDA$ is equal to

• A. ${\tan }^{-1}\frac{1}{19}$
• B. ${\tan }^{-1}\frac{2}{19}$
• C. ${\tan }^{-1}\frac{4}{19}$
• D. ${\tan }^{-1}\frac{8}{19}$

Answer 1

The correct option is B ${\tan }^{-1}\frac{2}{19}$


Parabola passing through $(0,0)$ and $(0,2)$
$k^2=-4h............\left(i\right)(2-k{)}^2=-4h......(ii)$
Using equation $\left(i\right)$ and $\left(ii\right)$,
$(2-k{)}^2=k^2\Rightarrow 4-4k+k^2=k^2\Rightarrow k=1$
Using equation$\left(i\right)$,
$h=-\frac{1}{4}$
Now,
$\tan \alpha =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|=\left|\frac{\left(\frac{k+2-k}{h+1-h}\right)-\left(\frac{k+2-1}{h+1-0}\right)}{1+\left(\frac{k+2-k}{h+1-h}\right)\times \left(\frac{k+2-1}{h+1-0}\right)}\right|=\left|\frac{2-\left(\frac{k+1}{h+1}\right)}{1+2\times \left(\frac{k+1}{h+1}\right)}\right|=\left|\frac{2-\left(\frac{8}{3}\right)}{1+2\times \left(\frac{8}{3}\right)}\right|=\frac{2}{19}$

Hence the $\angle PDA={\tan }^{-1}\frac{2}{19}$