SATHEE: Parabola 2 Question 13

Parabola 2 Question 13

13. The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the $X$ -axis is

(2001, 1M)

(a) $\sqrt{3} y = 3 x + 1$

(b) $\sqrt{3} y = - (x + 3)$

(d) $\sqrt{3} y = - (3 x + 1)$

Assertion and Reason

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Solution:

Any tangent to $y^{2} = 4 x$ is of the form $y = m x + \frac{1}{m}$ , $(∵ a = 1)$ and this touches the circle $(x - 3)^{2} + y^{2} = 9$ .

$| \frac{m (3) + \frac{1}{m} - 0}{\sqrt{m^{2} + 1}} | = 3$

$[∵$ centre of the circle is $(3, 0)$ and radius is 3 $]$ .

$\begin{array}{cc} \Rightarrow & \frac{3 m^{2} + 1}{m} = \pm 3 \sqrt{m^{2} + 1} \\ \Rightarrow & 3 m^{2} + 1 = \pm 3 m \sqrt{m^{2} + 1} \\ \Rightarrow & 9 m^{4} + 1 + 6 m^{2} = 9 m^{2} (m^{2} + 1) \\ \Rightarrow & 9 m^{4} + 1 + 6 m^{2} = 9 m^{4} + 9 m^{2} \\ \Rightarrow & 3 m^{2} = 1 \\ \Rightarrow & m = \pm \frac{1}{\sqrt{3}} \end{array}$

If the tangent touches the parabola and circle above the $X$ -axis, then slope $m$ should be positive.

$∴ m = \frac{1}{\sqrt{3}}$ and the equation is $y = \frac{1}{\sqrt{3}} x + \sqrt{3}$

$\sqrt{3} y = x + 3 .$

Parabola 2 Question 13

13. The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the $X$ -axis is

Assertion and Reason

Solution:

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Parabola 2 Question 13

13. The equation of the common tangent touching the circle (x−3)2+y2=9 and the parabola y2=4x above the X-axis is

Assertion and Reason

Solution:

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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13. The equation of the common tangent touching the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ above the $X$ -axis is