SATHEE: Parabola 1 Question 4

Parabola 1 Question 4

4. Let $P$ be the point on the parabola, $y^{2} = 8 x$ , which is at a minimum distance from the centre $C$ of the circle, $x^{2} + (y + 6)^{2} = 1$ . Then, the equation of the circle, passing through $C$ and having its centre at $P$ is

(2016 Main)

(a) $x^{2} + y^{2} - 4 x + 8 y + 12 = 0$

(b) $x^{2} + y^{2} - x + 4 y - 12 = 0$

(d) $x^{2} + y^{2} - 4 x + 9 y + 18 = 0$

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Answer:

Correct Answer: 4. (c)

Solution:

Centre of circle $x^{2} + (y + 6)^{2} = 1$ is $C (0, - 6)$ .

Let the coordinates of point $P$ be $(2 t^{2}, 4 t)$ .

Now, let

$\begin{aligned} D & = C P \\ = \sqrt{{(2 t^{2})}^{2} + (4 t + 6)^{2}} \end{aligned}$

$\Rightarrow$

Squaring on both sides

$\begin{array}{lrl} \Rightarrow & D^{2} (t) & = 4 t^{4} + 16 t^{2} + 48 t + 36 \\ Let & F (t) & = 4 t^{4} + 16 t^{2} + 48 t + 36 \\ For minimum, F^{'} (t) & = 0 \\ \Rightarrow 16 t^{3} + 32 t + 48 & = 0 \\ \Rightarrow & t^{3} + 2 t + 3 & = 0 \\ \Rightarrow & (t + 1) (t^{2} - t + 3) & = 0 \Rightarrow t = - 1 \end{array}$

Thus, coordinate of point $P$ are $(2, - 4)$ .

Now ,

$C P = \sqrt{2^{2} + (- 4 + 6)^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$

Hence, the required equation of circle is

$\begin{aligned} (x - 2)^{2} + (y + 4)^{2} & = (2 \sqrt{2})^{2} \\ \Rightarrow & x^{2} + 4 - 4 x + y^{2} + 16 + 8 y & = 8 \\ \Rightarrow & x^{2} + y^{2} - 4 x + 8 y + 12 & = 0 \end{aligned}$

Parabola 1 Question 4

4. Let $P$ be the point on the parabola, $y^{2} = 8 x$ , which is at a minimum distance from the centre $C$ of the circle, $x^{2} + (y + 6)^{2} = 1$ . Then, the equation of the circle, passing through $C$ and having its centre at $P$ is

Answer:

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Parabola 1 Question 4

4. Let P be the point on the parabola, y2=8x, which is at a minimum distance from the centre C of the circle, x2+(y+6)2=1. Then, the equation of the circle, passing through C and having its centre at P is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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, $y^{2} = 8 x$ , which is at a minimum distance from the centre $C$ of the circle, $x^{2} + (y + 6)^{2} = 1$ . Then, the equation of the circle, passing through $C$ and having its centre at $P$ is