SATHEE: Parabola 2 Question 3

Parabola 2 Question 3

3. The area (in sq units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the $X$ -axis is

(2019 Main, 9 April, II)

(a) $8 π (3 - 2 \sqrt{2})$

(b) $4 π (3 + \sqrt{2})$

(d) $4 π (2 - \sqrt{2})$

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Answer:

Correct Answer: 3. (c)

Solution:

Given parabola $y^{2} = 4 x$

So, equation of tangent to parabola (i) at point $(1, 2)$ is $2 y = 2 (x + 1)$

$[∵$ equation of the tangent to the parabola $y^{2} = 4 a x$ at a point $(x_{1}, y_{1})$ is given by $y y_{1} = 2 a (x + x_{1})]$

$\Rightarrow y = x + 1$

Now, equation of circle, touch the parabola at point $(1, 2)$ is

$\begin{matrix} (x - 1)^{2} + (y - 2)^{2} + λ (x - y + 1) = 0 \\ \Rightarrow x^{2} + y^{2} + (λ - 2) x + (- 4 - λ) y + (5 + λ) = 0 \end{matrix}$

Also, Circle (iii) touches the $x$ -axis, so $g^{2} = c$

$\begin{aligned} \frac{λ - 2}{2}^{2} = 5 + λ \\ \Rightarrow \begin{array}{c} 2 \\ λ^{2} - 4 λ + 4 = 4 λ + 20 \end{array} \\ \Rightarrow λ^{2} - 8 λ - 16 = 0 \\ \Rightarrow λ = \frac{8 \pm \sqrt{64 + 64}}{2} \\ \Rightarrow λ = 4 \pm \sqrt{32} = 4 \pm 4 \sqrt{2} \end{aligned}$

Now, radius of circle is $r = \sqrt{g^{2} + f^{2} - c}$

$\begin{aligned} r & = | f | \\ = \frac{λ + 4}{2} = \frac{8 + 4 \sqrt{2}}{2} or \frac{8 - 4 \sqrt{2}}{2} \end{aligned}$

For least area $r = \frac{8 - 4 \sqrt{2}}{2} = 4 - 2 \sqrt{2}$ units

So, area $= π r^{2} = π (16 + 8 - 16 \sqrt{2}) = 8 π (3 - 2 \sqrt{2})$ sq unit

Parabola 2 Question 3

3. The area (in sq units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the $X$ -axis is

Answer:

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Parabola 2 Question 3

3. The area (in sq units) of the smaller of the two circles that touch the parabola, y2=4x at the point (1,2) and the X-axis 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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3. The area (in sq units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the $X$ -axis is