Question

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 $P\left(\frac{1}{2},2\right)$ 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

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

The correct option is A $2$

Given equation of the parabola is $y^2=8x⟹a=2$

The endpoints of its latus rectum are $(a,2a)$ and $(a,-2a)$

Points are $(2,4)$ and $(2,-4)$

Area of the triangle ${\Delta }_1=\left|\begin{array}{ccccc}0.5& & 2& & 1\\ 2& & 4& & 1\\ 2& & -4& & 1\end{array}\right|=6$

The equation of the tangent at $(2,4)$ is given by $y=x+2\dots \left(1\right)$

equation of the tangent at $(2,-4)$ is given by $-y=x+2\dots \left(2\right)$

and the equation of the tangent at $(0.5,2)$ is given by $y=2x+1\dots \left(3\right)$

Intersection points of $\left(1\right),\left(2\right)$ and $\left(3\right)$ are : $(-2,0),(-1,-1)$ and $(1,3)$

${\Delta }_2=\left|\begin{array}{ccccc}-2& & 1& & 1\\ -1& & -1& & 1\\ 1& & 3& & 1\end{array}\right|=3$

$\therefore \frac{{\Delta }_1}{{\Delta }_2}=2$