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

• A

  $\tan ^{-1} \frac{1}{19}$
• Correct answer
  B

  $\tan ^{-1} \frac{2}{19}$
• C

  $\tan ^{-1} \frac{4}{19}$
• D

  $\tan ^{-1} \frac{8}{19}$

Solution

Curve, $S:(y-k)^{2}=4(x-h)$ LLR $=4 ;$ Clearly $\mathrm{k}=1 ; \Rightarrow \mathrm{A}(\mathrm{h}, 1) \& ' \mathrm{M}^{\prime}$ is focus $(\mathrm{h}+1,1)$ So D $(\mathrm{h}+1,3)$ $\mathrm{S}_{(0,0)}=0 \Rightarrow \mathrm{k}^{2}=-4 \mathrm{~h}$ $\Rightarrow \mathrm{h}=\frac{-1}{4} \quad \Rightarrow \mathrm{D}\left(\frac{3}{4}, 3\right)$ Now $; \tan \alpha=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\left|\frac{\frac{8}{3}-2}{1+\frac{8}{3} \times 2}\right|=\frac{2}{19}$ where, $m_{1}=\frac{3-1}{\frac{3}{4}-0}=\frac{2}{\frac{3}{4}}=\frac{8}{3}$ $$ \mathrm{m}_{2}=\frac{3-1}{1}=2 $$

Hence, the correct answer is option: 'B'.