ParabolaMedium

Question

Let the directrix of the parabola $P:y^{2} = 8x$, cut x -axis at the point A . Let $B(\alpha,\beta),\alpha > 1$, be a point on P such that the slope of AB is $3/5$. If BC is a focal chord of P , then six times the area of $\bigtriangleup ABC$ is:

Options

A.80
B.160
C.174
D.192

Solution

Sol. [Key Concept: Parametrise $B = (2t^{2},4t)$; the slope of $AB = \frac{3}{5}$ fixes $t$, C is its focal-chord partner, then compute $6 \times$ area of $triangle\ ABC$.]

Let $B\left( 2t^{2},4t \right) \Rightarrow C\left( \frac{2}{t^{2}},\frac{- 4}{t} \right);A( - 2,0)$

given $m_{AB} = \frac{4t}{2t^{2} + 2} = \frac{3}{5}$

$${20t = 6t^{2} + 6 }{3t^{2} - 10t + 3 = 0 }$$$t = 3,\frac{1}{3}$ but $t > 1$ (given)

$$\Rightarrow P(18,12),Q\left( \frac{2}{9},\frac{- 4}{3} \right)$$

$$ {A = \left| \frac{1}{2} \right|\begin{matrix} - 2 & 0 & 1 \\ 18 & 12 & 1 \\ \frac{2}{9} & \frac{- 4}{3} & 1 \end{matrix}|| = \frac{80}{3}}$$

$$ {\Rightarrow 6\text{ }A = 160 }$$

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