Parabola 2 Question 24
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) $8 \pi(3-2 \sqrt{2})$
(b) $4 \pi(3+\sqrt{2})$
(c) $8 \pi(2-\sqrt{2})$
(d) $4 \pi(2-\sqrt{2})$
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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 $2 y=2(x+1)$
$\left[\because\right.$ equation of the tangent to the parabola $y^{2}=4 a x$ at a point $\left(x _1, y _1\right)$ is given by $\left.y y _1=2 a\left(x+x _1\right)\right]$
$$ \Rightarrow \quad y=x+1 $$
Now, equation of circle, touch the parabola at point $(1,2)$ is
$$ \begin{gathered} (x-1)^{2}+(y-2)^{2}+\lambda(x-y+1)=0 \\ \Rightarrow \quad x^{2}+y^{2}+(\lambda-2) x+(-4-\lambda) y+(5+\lambda)=0 \end{gathered} $$
Also, Circle (iii) touches the $x$-axis, so $g^{2}=c$
$$ \begin{aligned} & \frac{\lambda-2}{2}{ }^{2}=5+\lambda \\ & \Rightarrow \quad \begin{array}{c} 2 \\ \quad \lambda^{2}-4 \lambda+4=4 \lambda+20 \end{array} \\ & \Rightarrow \quad \lambda^{2}-8 \lambda-16=0 \\ & \Rightarrow \quad \lambda=\frac{8 \pm \sqrt{64+64}}{2} \\ & \Rightarrow \quad \lambda=4 \pm \sqrt{32}=4 \pm 4 \sqrt{2} \end{aligned} $$
Now, radius of circle is $r=\sqrt{g^{2}+f^{2}-c}$
$$ \begin{aligned} r & =|f| \\ & =\frac{\lambda+4}{2}=\frac{8+4 \sqrt{2}}{2} \text { or } \frac{8-4 \sqrt{2}}{2} \end{aligned} $$
For least area $r=\frac{8-4 \sqrt{2}}{2}=4-2 \sqrt{2}$ units
So, area $=\pi r^{2}=\pi(16+8-16 \sqrt{2})=8 \pi(3-2 \sqrt{2})$ sq unit
