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 $=\frac{3-1}{\frac{3}{4}+\frac{1}{4}}-2-m_2$
$\therefore \angle PDA=\theta ={\tan }^{-1}\left(\frac{m_1-m_2}{1+m_1{m}_2}\right)$
$={\tan }^{-1}\left(\frac{8/3-2}{1+\frac{16}{3}}\right)$
$={\tan }^{-1}\left(\frac{2}{19}\right)$