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 latusrectum. 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

Answer: (B)

Given parabola $(y-k)^{2}=4(x-h)$
This is passes through $(0, 0)$ and $(0, 2)$.
$\therefore k^{2}=-4h \dots(i)$
and $(2-k)^{2}=-4h \dots(ii)$
From Eqs. (i) and (ii), we get
$k=1, h=-\frac{1}{4}$
$\therefore $ Equation of parabola becomes
$(y-1)^{2}=4x+1$
Focus of parabola = $\left(\frac{3}{4}, 1\right)$
End of latusrectum $D=\left(\frac{3}{4}, 3\right)$
Axis of parabola is $y - 1 = 0$
Point P intersection of axis of parabola and Y-axis.
$\therefore P(0, 1)$
$D=\left(\frac{3}{4}, 3\right)$
$A\left(-\frac{1}{4}, 1\right)$ $[\because$ A is vertex of parabola ]
Slope of PD $=\frac{3-1}{3 /4 -0}=8 /3=m_{1}$
Slope of DA $=\frac{3-1}{\frac{3}{4}+\frac{1}{4}}-2-m_{2}$
$\therefore \angle\, PDA =\theta=\tan^{-1} \left(\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right)$
$=\tan^{-1}\left(\frac{8/ 3-2}{1+\frac{16}{3}}\right)$
$=\tan^{-1}\left(\frac{2}{19}\right)$