Question

Let $A(1,2),B(4,-4),C(2,2\sqrt{2})$ be points on the parabola $y^2=4x$. If $\alpha$ and $\beta$ respectively represent the area of $△ABC$ 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, $y^2=4x$
$\Rightarrow 2y\frac{dy}{dx}=4$
$\Rightarrow \frac{dy}{dx}=\frac{2}{y}$
At $A(1,2)$
${\frac{dy}{dx}|}_{(1,2)}=\frac{2}{2}=1$
$\therefore$ Tangent is given by
$y-2=1(x-1)$
$y-2=x-1$
$y=x+1\dots \dots$ (i)
At $B(4,-4)$
${\left(\frac{dy}{dx}\right)}_{(4,-4)}=\frac{2}{-4}=-\frac{1}{2}$
$\therefore$ Tangent is given by $y+4=\frac{-1}{2}(x-4)$
$2y+8=-x+4$
$x+2y+4=0\dots \dots$(ii)
At $C(2,2\sqrt{2})$
${\frac{dy}{dx}|}_{(2,2\sqrt{2})}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\therefore$ 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, $ar(\Delta ABC)=\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 4& -4& 1\\ 2& 2\sqrt{2}& 1\end{array}\right|$
$=\frac{1}{2}[1(-4-2\sqrt{2})-2(4-2)+1(8\sqrt{2}+8)]$
$=\frac{1}{2}[-4-2\sqrt{2}-4+8\sqrt{2}+8]=3\sqrt{2}$ sq unit
and $ar$
$(\Delta PQR)=\frac{1}{2}\left|\begin{array}{ccc}-2& -1& 1\\ -2\sqrt{2}& \sqrt{2}-2& 1\\ \sqrt{2}& \sqrt{2}+1& 1\end{array}\right|$
$=\frac{1}{2}[-2[\sqrt{2}-2-\sqrt{2}-1]+1[-2\sqrt{2}-\sqrt{2}]+1[-4-2\sqrt{2}-2+2\sqrt{2}]]$
$=\frac{1}{2}[6-3\sqrt{2}-6]=\frac{-3}{2}\sqrt{2}$
$=\frac{3}{2}\sqrt{2}sq$ [area cannot be negative]
$\therefore \alpha =3\sqrt{2}$ and $\beta =\frac{3}{2}\sqrt{2}$
$\therefore \alpha \beta =3\sqrt{2}\times \frac{3}{2}\sqrt{2}=9$