JEE Advanced 2011 Paper 1Question 1
Consider the parabola $y^2=8x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its latus rectum and the point $P\left(\frac12,2\right)$ on the parabola, and $\Delta_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latus rectum. Then $\frac{\Delta_1}{\Delta_2}$ is
Parabola
Answer$2$
Solution
Formulas used. (1) Tangent to $y^2=4ax$ at $(x_1,y_1)$: $yy_1=2a(x+x_1)$(2) Area $=\frac12\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big|$Step 1 - Latus rectum. $y^2=8x$, $a=2$: ends $L(2,4)$, $L'(2,-4)$. Step 2 - $\Delta_1$. Base $LL'=8$ on $x=2$; height = distance of $P$ from $x=2$ $=2-\frac12=\frac32$: $$\Delta_1=\frac12\cdot8\cdot\frac32=6$$ Step 3 - Tangents, use (1). At $P\big(\frac12,2\big)$: $2y=4\big(x+\frac12\big)\Rightarrow y=2x+1$ At $L(2,4)$: $4y=4(x+2)\Rightarrow y=x+2$ At $L'(2,-4)$: $-4y=4(x+2)\Rightarrow y=-x-2$ Step 4 - Vertices. $2x+1=x+2\Rightarrow(1,3)$; $2x+1=-x-2\Rightarrow(-1,-1)$; $x+2=-x-2\Rightarrow(-2,0)$.Step 5 - $\Delta_2$, use (2). $$\Delta_2=\frac12\big|1(-1-0)+(-1)(0-3)+(-2)(3+1)\big|=\frac12|-1+3-8|=3$$ Step 6. $\frac{\Delta_1}{\Delta_2}=\frac63=2$. Answer: 2