Question

The intercept of the latus rectum to the parabola $y^2=4ax$ are $b$ and $k$, then $k$ is equal to:

• A. $\frac{ab}{a-b}$
• B. $\frac{a}{b-a}$
• C. $\frac{b}{b-a}$
• D. $\frac{ab}{b-a}$

Answer 1

The correct option is D $\frac{ab}{b-a}$

For latusrectum $PQ$,
$y(t_1+t_2)-2x-2a{t}_1{t}_2=0$
and $t_1{t}_2=-1$
For any point $P(x,y)$, the focal distance is $a+x$.
$\therefore b=a+x=a+a{t}_1^2=a(1+t_1^2)$ ...... (i)
$c=a+x=a+a{t}_2^2=a+\frac{a}{t_1^2}(\because t_1{t}_2=-1)$
$=\frac{a(1+t_1^2)}{t_1^2}$ ..... (ii)
$\therefore \frac{b}{c}=t_1^2$
$\therefore$ From Eq. (i),
$b=a(1+\frac{b}{c})$
$\Rightarrow b=a+\frac{ab}{c}$
$\therefore c=\frac{ab}{b-a}$