Test Index

Parabola Part 1

Section: Mathematics

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Question : 68 of 100

Marks: +1, -0

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

[9 Apr 2019 Shift 2]

4 $\pi(2-\sqrt{2})$
4 $\pi(3+\sqrt{2})$
8 $\pi(2-\sqrt{2})$
8 $\pi(3-2\sqrt{2})$

Solution:

Given parabola

\Rightarrow y^{2}=4 x

\Rightarrow

Let centre is

(x_{1}, y_{1})

(x-x_{1})^{2}+(y-y_{1})^{2}=y_{1}^{2}

is the eq of circele

x_{1}^{2}-2 x_{1}-4 y_{1}+5=0 ......

(1)

\Rightarrow

normal at

(1,2)

to the parabola is x+y = 3

\Rightarrow x_{1}+y_{1}=3 ........(2)(

It passes through centre

)

\Rightarrow x_{1}= \frac{-2 \pm \sqrt{4+4(7)}}{2}

=-1+2\sqrt{2}

=2\sqrt{2}-1

area

=\pi r^{2}=\pi y_{1}^{2}

=\pi[16+8-16\sqrt{2}]

=8\pi(3-2\sqrt{2})

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