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 ${\tan }^{-1}\frac{2}{19}$

Given Parabola: $({y-k)}^2=4(x-h).............(i)$

Vertex $A$:$(h,k)$

Focus $S$:$(1+h,k)$

Passes through $O(0,0)$ $=>(0+k{)}^2=4(0-h)$

$=>k^2=-4h...............\left(ii\right)$

Also Passes through $\angle (0,2)=>(2-k{)}^2=4(0-h).................(iii)$

From $\left(ii\right)$and$\left(iii\right)$, $k^2=k^2+4-4k$

$=>k=1$ and $h=\frac{-1}{4}...................\left(iv\right)$

End point of $\angle R:(1-\frac{1}{4},1\pm 2),D:(\frac{3}{4},3)and(\frac{3}{4},-1)$

Axis of parabola: $y-k=0=>y=\mathrm{1............}\left(v\right)$

Point of intersection of $y=1$ with $y-axis$ ie. $x=0$ is $P:(0,1)$

Line Passing through $PD$:$(y-1)=\left(\frac{3-1}{\frac{3}{4}}\right)(x-0)$

$:y=\frac{8}{3}x+1$

Slope of $PD$, $m=\frac{8}{3}......................\left(vi\right)$

Line Passing through $AD$, $(y-1)=\left(\frac{3-1}{\frac{3}{4}+\frac{1}{4}}\right)(x+\frac{1}{4})$

$=>y=2x+\frac{1}{2}+1$

Slope of $AD=m_2=\mathrm{2..........}\left(vii\right)$

Angle between $AD$ and $PD=>\angle PDA={tan}^{-1}\left|\frac{\frac{8}{3}-2}{1+\frac{8}{3}\left(2\right)}\right|$

$\angle PDA={tan}^{-1}\left|\frac{\frac{2}{3}}{\frac{19}{3}}\right|$

$\angle PDA={tan}^{-1}\left(\frac{2}{19}\right)$.