Parabola 3 Question 1
1. If the parabolas $y^{2}=4 b(x-c)$ and $y^{2}=8 a x$ have a common normal, then which one of the following is a valid choice for the ordered triad $(a, b, c)$ ?
(2019 Main, 10 Jan, I)
(a) $\frac{1}{2}, 2,0$
(b) $(1,1,0)$
(c) $(1,1,3)$
(d) $\frac{1}{2}, 2,3$
Show Answer
Answer:
Correct Answer: 1. (c)
Solution:
- 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 $m=\tan \theta=\frac{y _2-y _1}{x _2-x _1}$.
$\therefore$ 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 $=4 a \operatorname{cosec}^{2} \alpha$
$\therefore$ The required length of the focal chord
$$ \begin{aligned} & =16\left[1+\cot ^{2} \alpha\right] \quad\left[\because a=4 \text { and } \operatorname{cosec}^{2} \alpha=1+\cot ^{2} \alpha\right] \\ & =161+\frac{9}{16}=25 \text { units } \quad \because \cot \alpha=\frac{1}{\tan \alpha}=-\frac{3}{4} \end{aligned} $$
