SATHEE: Parabola 1 Question 2

Parabola 1 Question 2

2. A circle cuts a chord of length $4 a$ on the $X$ -axis and passes through a point on the $Y$ -axis, distant $2 b$ from the origin. Then, the locus of the centre of this circle, is

(2019 Main, 11 Jan, II)

(a) a parabola

(b) an ellipse

(d) a hyperbola

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Answer:

Correct Answer: 2. (a)

Solution:

According to given information, we have the following figure.

Let the equation of circle be

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

According the problem,

$4 a = 2 \sqrt{g^{2} - c}$

$[∵$ The length of intercepts made by the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

$with X -axis is 2 \sqrt{g^{2} - c}]$

Also, as the circle is passing through $P (0, 2 b)$

$\begin{array}{lrr} ∴ & 0 + 4 b^{2} + 0 + 4 b f + c = 0 & [using Eq. (i)] \\ \Rightarrow & 4 b^{2} + 4 b f + c = 0 & \dots (iii) \end{array}$

Eliminating ’ $c$ ’ from Eqs. (ii) and (iii), we get

$\begin{aligned} 4 b^{2} + 4 b f + g^{2} - 4 a^{2} = 0 \\ [∵ 4 a = 2 \sqrt{g^{2} - c} \Rightarrow c = g^{2} - 4 a^{2}] \end{aligned}$

So, locus of $(- g, - f)$ is

$\begin{aligned} 4 b^{2} - 4 b y + x^{2} - 4 a^{2} = 0 \\ \Rightarrow & x^{2} = 4 b y + 4 a^{2} - 4 b^{2} \end{aligned}$

which is a parabola.

Parabola 1 Question 2

2. A circle cuts a chord of length $4 a$ on the $X$ -axis and passes through a point on the $Y$ -axis, distant $2 b$ from the origin. Then, the locus of the centre of this circle, is

Answer:

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Parabola 1 Question 2

2. A circle cuts a chord of length 4a on the X-axis and passes through a point on the Y-axis, distant 2b from the origin. Then, the locus of the centre of this circle, 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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2. A circle cuts a chord of length $4 a$ on the $X$ -axis and passes through a point on the $Y$ -axis, distant $2 b$ from the origin. Then, the locus of the centre of this circle, is