SATHEE: Parabola 3 Question 4

Parabola 3 Question 4

4. 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) $S Q : Q P = (\sqrt{5} + 1) : 2$

(d) the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

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Answer:

Correct Answer: 4. (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 $A B$ .

$M P = t_{1} + t_{2} = r$

Also, $m_{A B} = \frac{2 t_{2} - 2 t_{1}}{t_{2}^{2} - t_{1}^{2}} = \frac{2}{t_{2} + t_{1}}$ [when $A B$ is chord]

$\begin{array}{lrr} \Rightarrow & m_{A B} = \frac{2}{r} & [from Eq. (i)] \\ Also, & m_{A^{'} B^{'}} = - \frac{2}{r} & [when A^{'} B^{'} is chord] \end{array}$

Hence, (c, d) are the correct options.

Parabola 3 Question 4

4. 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,

Answer:

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Parabola 3 Question 4

4. Let P be the point on the parabola y2=4x, which is at the shortest distance from the centre S of the circle x2+y2−4x−16y+64=0. Let Q be the point on the circle dividing the line segment SP internally. Then,

Answer:

Engineering Preparation

Medical Preparation

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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4. 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,