Parabola Ques 53
- Let $P$ be the point on the parabola $y^{2}=4 x$, which is at the shortest distance from the centre $S$ of the circle $x^{2}+y^{2}-4 x-16 y+64=0$. Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then,
(a) $S P=2 \sqrt{5}$
(2016 Adv.)
(b) $S Q: Q P=(\sqrt{5}+1)/2$
(c) the $x$-intercept of the normal to the parabola at $P$ is 6
(d) the slope of the tangent to the circle at $Q$ is $- \frac{1}{2}$
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Answer:
Correct Answer: 53.(a, c, d)
Solution:
- Here, coordinate $M=\frac{t_1^{2}+t_2^{2}}{2}, t_1+t_2$ i.e. mid-point of chord $A B$.
$$ M P = t₁ + t₂ = r $$
Also, $\quad m _{A B}=\frac{2 t _2-2 t _1}{t _2^{2}-t _1^{2}}=\frac{2}{t _2+t _1} \quad$ [when $A B$ is chord]
$$ \begin{array}{lrr} \Rightarrow & m _{A B}=\frac{2}{r} & \text { [from Eq. (i)] } \\ \text { Also, } & m _{A^{\prime} B^{\prime}}=-\frac{2}{r} & \text { [when } A^{\prime} B^{\prime} \text { is chord] } \end{array} $$
Hence, (c, d) are the correct options.
BETA
