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

R.E.F image

Given a parabola $(y-k{)}^2=4(x-h)...(1)$

Since, the given parabola passes through $O=(0,0)$

$\therefore$ the given O must satisfy the $e{q}^n$ of parabola

So, $(O-k{)}^2=4(O-h)$

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

Similarly, Pint $L=(0,2)$ must satisfy the $e{q}^n$ of parabola

$\therefore (2-k{)}^2=4(0-h)$

$\Rightarrow 4-4k+k^2=-4h$

$\Rightarrow 4-4k+k^2=k^2$ (By $e{q}^n\left(2\right))$

$\Rightarrow 4-4k+k^2=0$

$\Rightarrow 4-4k=0$

$\Rightarrow k=1$

Substituting in $e{q}^n$ (2), we get

$1=-4h$

$\Rightarrow h=\frac{-1}{4}$

Hence, the $e{q}^n$ of parabola is

$(y-1{)}^2=4(x+\frac{1}{4})...\left(3\right)$

Now, Put $X=x+\frac{1}{4}$ & $Y=y-1$

Then, $e{q}^n\left(3\right)$ becomes

$Y^2=4x...\left(4\right)$

Vertex of the parabola

$X=0$ $Y=0$

$x+\frac{1}{4}=0$ $y-1=0$

$x=\frac{-1}{4}$ $y=1$

$\therefore$ Vertex $A=(\frac{-1}{4},1)$

Axis of the parabola

$y=0$

$\Rightarrow y-1=0$

$\Rightarrow y=1$

Foues of the parabola

$X=a$ $Y=0$

from $e{q}^2=4x$

on comparing with $y^2=4ax$ we get

$4a=4$

$a=1$

So, $x=1$ $Y=0$

$\Rightarrow x+\frac{1}{4}=1$ $y-1=0$

$\Rightarrow x=1-\frac{1}{4}$ $y=1$

$\Rightarrow x=3/4$

foues $S=(\frac{3}{4},1)$

Length of Latus Rectum

length of latus Rectum $D{D}^1=4a=4\times =4$

Now, let the coordinate of the point D is $(x,y)$

By the symmetry of the parabola $SD=\frac{D{D}^1}{2}$

$\Rightarrow SD=\frac{4}{2}=2$

$\Rightarrow (SD{)}^2=4$ (On squaring both sides)

$\Rightarrow (x_1-\frac{3}{4}{)}^2+(y_1-1{)}^2=4...\left(5\right)$ (By using distance formula)

$\because SD$ is perpendicular to the axis of the parabola $y=1$

$\therefore$ distance of the point D from the line $y-1=0$ Ps

$|\frac{ox{x}_1+1x{y}_1-1}{\sqrt{o^2+1^2}}|=SD=2$

$\Rightarrow y_1-1=2$

$\Rightarrow y_1=3$

Substituing in $e{q}^n\left(5\right)$, we get $(x,\frac{-3}{4}{)}^2+(2{)}^2=4\Rightarrow (x,\frac{3}{4}{)}^2+4=4$

$\Rightarrow (x_1-\frac{3}{4}{)}^2=0$

$\Rightarrow x_1=3/4$

$\therefore$ Coordinate of point D is $(\frac{3}{4},3)$

Now, $\angle PDA=\theta =$ Angle b/w the two lines AD & PD

Slope of line AD $m_1=\frac{3-1}{\frac{3}{4}+\frac{1}{4}}=\frac{2}{1}=2$

Slope of line PD $m_2=\frac{3-1}{\frac{3}{4}-0}=\frac{2}{\frac{3}{4}}=\frac{8}{3}|$

$\therefore \angle PDA=ta{n}^{-1}|\frac{m_1-m_2}{1+m_1{m}_2}|=ta{n}^{-1}|\frac{2-8/3}{1+2\times \frac{8}{3}}|$

$\angle PDA=ta{n}^{-1}|\frac{2-8/3}{1+2\times \frac{8}{3}}|$

$\angle PDA=ta{n}^1|\frac{\frac{6-8}{3}}{\frac{3+16}{3}}|=ta{n}^{-1}|\frac{-2}{3}\times \frac{3}{19}|=ta{n}^{-1}|\frac{-2}{19}|$

$\therefore \angle PDA=ta{n}^{-1}\left(\frac{2}{19}\right)$