Question

Let $A(1,2), B(4,-4), C(2,2 \sqrt{2})$ be points on the parabola $y^{2}=4 x$. If $\alpha$ and $\beta$ respectively represent the area of $\triangle ABC$ and the area of the triangle formed by the tangents at $A, B, C$ to the above parabola, then $\alpha \beta=$

• A. $6$
• B. $3 \sqrt{2}$
• C. $9$
• D. $6 \sqrt{2}$

Answer 1

Answer: (C)

We have, $y^{2}=4 x$
$\Rightarrow 2 y \frac{d y}{d x}=4 $
$\Rightarrow \frac{d y}{d x}=\frac{2}{y}$
At $A(1,2)$
$\left.\frac{d y}{d x}\right|_{(1,2)}=\frac{2}{2}=1$
$\therefore $ Tangent is given by
$y-2=1(x-1)$
$y-2=x-1$
$y=x+1 \ldots \ldots$ (i)
At $B(4,-4)$
$\left(\frac{d y}{d x}\right)_{(4,-4)}=\frac{2}{-4}=-\frac{1}{2}$
$\therefore $ Tangent is given by $y+4=\frac{-1}{2}(x-4)$
$2 y+8=-x+4$
$x+2 y+4=0 \ldots \ldots$(ii)
At $C(2,2 \sqrt{2})$
$\left.\frac{d y}{d x}\right|_{(2,2 \sqrt{2})}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}$
$\therefore $ Tangent is given by
$y-2 \sqrt{2}=\frac{1}{\sqrt{2}}(x-2)$
$\Rightarrow \sqrt{2} y-4=x-2$
$\Rightarrow x-\sqrt{2} y+2=0$.......(iii)
Let $P, Q, R$ the vertices of triangle formed by tangents (i), (ii), (iii).
On solving Eqs. (i), (ii) and (iii), we get
$P(-2,-1), Q(-2 \sqrt{2}, \sqrt{2}-2)$ and $R(\sqrt{2}, \sqrt{2}+1)$
Now, $ar (\Delta A B C)=\frac{1}{2}\begin{vmatrix}1 & 2 & 1 \\ 4 & -4 & 1 \\ 2 & 2 \sqrt{2} & 1\end{vmatrix}$
$=\frac{1}{2}[1(-4-2 \sqrt{2})-2(4-2)+1(8 \sqrt{2}+8)] $
$=\frac{1}{2}[-4-2 \sqrt{2}-4+8 \sqrt{2}+8]=3 \sqrt{2}$ sq unit
and $ar$
$(\Delta P Q R)=\frac{1}{2}\begin{vmatrix}-2 & -1 & 1 \\ -2 \sqrt{2} & \sqrt{2}-2 & 1 \\ \sqrt{2} & \sqrt{2}+1 & 1\end{vmatrix}$
$=\frac{1}{2}[-2[\sqrt{2}-2-\sqrt{2}-1]+1[-2 \sqrt{2}-\sqrt{2}]+1[-4-2 \sqrt{2}-2+2 \sqrt{2}]]$
$=\frac{1}{2}[6-3 \sqrt{2}-6]=\frac{-3}{2} \sqrt{2}$
$=\frac{3}{2} \sqrt{2} s q$ [area cannot be negative]
$\therefore \alpha=3 \sqrt{2}$ and $\beta=\frac{3}{2} \sqrt{2}$
$\therefore \alpha \beta=3 \sqrt{2} \times \frac{3}{2} \sqrt{2}=9$