Parabola Ques 1
- If one end of a focal chord of the parabola, $y^2=16 x$ is at $(1,4)$, then the length of this focal chord is
(2019 Main, 9 April, 1)
(a) $22$
(b) $25$
(c) $24$
(d) $20$
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Answer:
Correct Answer: 1.(b)
Solution: (b) Key Idea
(i) First find the focus of the given parabola
(ii) Then, find the slope of the focal chord by using $m=\frac{y_2-y_1}{x_2-x_1}$
(iii) Now, find the length of the focal chord by using the formula $4 a \operatorname{cosec}^2 \alpha$
Equation of given parabola is $y^2=16 x$, its focus is $(4,0)$.
Since, slope of line passing through $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by $m=\tan \theta=\frac{y_2-y_1}{x_2-x_1}$.
$\therefore \quad $ Slope of focal chord having one end point is $(1,4)$ is
$ m=\tan \alpha=\frac{4-0}{1-4}=-\frac{4}{3} $
[where, ’ $\alpha$ ’ is the inclination of focal chord with $X$-axis.] Since, the length of focal chord $=4 a \operatorname{cosec}{ }^2 \alpha$
$\therefore \quad $ The required length of the focal chord
$ \begin{aligned} & =16\left[1+\cot ^2 \alpha\right] \quad\left[\because a=4 \text { and } \operatorname{cosec}^2 \alpha=1+\cot ^2 \alpha\right] \\ & =16\left[1+\frac{9}{16}\right]=25 \text { units } \quad\left[\because \cot \alpha=\frac{1}{\tan \alpha}=-\frac{3}{4}\right] \end{aligned} $
BETA
