Q89 / 150 · MHT CET 2012 · Mathematics

The intercept of the latusrectum to the parabola $y^{2}=4 a x$ are $b$ and $k$, then $k$ is equal to

  • A
    $\frac{a b}{a-b}$
  • B
    $\frac{a}{b-a}$
  • C
    $\frac{b}{b-a}$
  • Correct answer
    D
    $\frac{a b}{b-a}$

Solution

For latusrectum $P Q$. $$ y\left(t_{1}+t_{2}\right)-2 x-2 a t_{1} t_{2}=0 $$ and $\quad t_{1} t_{2}=-1$ For any point $P(x, y)$, the focal distance is $a+x$. $\therefore \quad b=a+x=a+a t_{1}^{2}=a\left(1+t_{1}^{2}\right) \quad \ldots(\mathrm{i})$ $c=a+x=a+a t_{2}^{2}=a+\frac{a}{t_{1}^{2}} \quad\left(\because t_{1} t_{2}=-1\right)$ $=\frac{a\left(1+t_{1}^{2}\right)}{t_{1}^{2}}$...(ii) $\frac{b}{c}=t_{1}^{2}$ $\therefore$ From Eq. (i), $b=a\left(1+\frac{b}{c}\right)$ $\Rightarrow \quad b=a+\frac{a b}{c}$ $c=\frac{a b}{b-a}$

Hence, the correct answer is option: 'D'.

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