Search any question & find its solution
Question:
Answered & Verified by Expert
The intercept of the latusrectum to the parabola $y^{2}=4 a x$ are $b$ and $k$, then $k$ is equal to
Options:
Solution:
2252 Upvotes
Verified Answer
The correct answer is:
$\frac{a b}{b-a}$
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), $b=a\left(1+\frac{b}{c}\right)$
$\Rightarrow \quad b=a+\frac{a b}{c}$
$c=\frac{a b}{b-a}$
$$
y\left(t_{1}+t_{2}\right)-2 x-2 a t_{1} t_{2}=0
$$
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), $b=a\left(1+\frac{b}{c}\right)$
$\Rightarrow \quad b=a+\frac{a b}{c}$
$c=\frac{a b}{b-a}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.