Question

$AB$ is a double ordinate of the parabola $y^2=4ax$. Tangents drawn to parabola at $A$ and $B$ meet $y-$axis at $A_1$ and $B_1$, respectively. If the area of trapezium $A{A}_1{B}_{1}B$ is equal to $12a^2$, then angle subtended by $A_1{B}_1$ at the focus of the parabola is equal to

• A. $2{\tan }^{-1}\left(3\right)$
• B. ${\tan }^{-1}\left(3\right)$
• C. $2{\tan }^{-1}\left(2\right)$
• D. ${\tan }^{-1}\left(2\right)$

Answer 1

Answer: (C)

$2{\tan }^{-1}\left(2\right)$

Let $A\equiv (a{t}_1^2,2a{t}_1),B\equiv (a{t}_1^2,-2a{t}_1)$ are points on the parabola $y^2=4ax$
Slope of parabola $=\frac{dy}{dx}=\frac{2a}{y}$
Slope of tangent at $A=\frac{1}{t_1}$
Equation of tangent at $A$
$y-2a{t}_1=\frac{1}{t_1}(x-a{t}_1^2)$
$\Rightarrow t_{1}y=x+a{t}_1^2$
Slope of tangent at $B=-\frac{1}{t_1}$
Equation of tangent at $B$
$y+2a{t}_1=-\frac{1}{t_1}(x-a{t}_1^2)$
$\Rightarrow -t_{1}y=x+a{t}_1^2$
Since, these tangents meet at y-axis i.e $x=0$
So the points at which tangents meet at y-axis are $A_1=(0,a{t}_1),B_1=(0,-a{t}_1)$
Area of trapezium $A{A}_1{B}_{1}B=\frac{1}{2}(AB+A_1{B}_1)\times OC$
$12a^2=\frac{1}{2}(4a{t}_1+2a{t}_1)a{t}_1^2$
$\Rightarrow t_1=2$
Let $\angle OS{A}_1=\theta$
In triangle $OS{A}_1$,
$\tan \theta =\frac{2a}{a}$
$\Rightarrow \theta ={\tan }^{-1}\left(2\right)$
Hence, the angle subtended by $A_1{B}_1$ at the focus of the parabola $=2{\tan }^{-1}\left(2\right)$