Parabola Ques 43
- Equation of a common tangent to the circle, $x^{2}+y^{2}-6 x=0$ and the parabola, $y^{2}=4 x$, is
(2019 Main, 9 Jan, I)
(a) $\sqrt{3} y=3 x+1$
(b) $2 \sqrt{3} y=12 x+1$
(c) $\sqrt{3} y=x+3$
(d) $2 \sqrt{3} y=-x-12$
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Answer:
Correct Answer: 43.(d)
Solution:
- We know that, equation of tangent to parabola $y^{2}=4 a x$ is $y = mx + \frac{a}{m}$
$$ y=m x+\frac{a}{m} $$
$\therefore$ Equation of tangent to the parabola $y^{2}=4 x$ is $y = mx + \frac{1}{m}$
$$ \begin{alignedat} & y=m x+\frac{1}{m} \\ \Rightarrow \quad & m^{2} x-m y+1=0 \end{aligned} $$
Now, let line (i) is also a tangent to the circle.
Equation of circle $x^{2}+y^{2}-6 x=0$
Clearly, centre of given circle is $(3,0)$ and radius $=3$
$\left[\because\right.$ for the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$, centre $=(-g,-f)$
$$ (-g,-f) \text { and radius }=\sqrt{g^{2}+f^{2}-c} $$
$\therefore$ The perpendicular distance of $(3,0)$ from the line (i) is 3.
$[\because$ Radius is perpendicular to the tangent line at the point of contact
$\Rightarrow \quad \frac{\left|m^{2} \cdot 3-m \cdot 0+1\right|}{\sqrt{\left(m^{2}\right)^{2}+(-m)^{2}}}=3$
tangent of a circle]
The length of perpendicular from a point $\left(x _1, y _1\right)$ to the line $a x+b y+c=0$ is $\left|\frac{a x _1+b y _1+c}{\sqrt{a^{2}+b^{2}}}\right|$.
$\Rightarrow \quad \frac{3 m^{2}+1}{\sqrt{m^{4}+m^{2}}}\neq 3$
$\Rightarrow 9 m^{4}+6 m^{2}+1=9\left(m^{4}+m^{2}\right)+1$
$\Rightarrow \quad m \approx \infty$ or $m= \pm \frac{1}{\sqrt{3}}$
$$ \because \lim _{m \rightarrow \infty} \frac{3 m^{2}+1}{\sqrt{m^{4}+m^{2}}}=\lim _{m \rightarrow \infty} \frac{3+\frac{1}{m^{2}}}{\sqrt{1+\frac{1}{m^{2}}}}=3 $$
$\therefore$ Equation of common tangents are $x=0$ and $y=0$,
$$ y=\frac{x}{\sqrt{3}}+\sqrt{3} \text { and } y=\frac{-x}{\sqrt{3}}-\sqrt{3} \quad \text { using } y=m x+\frac{1}{m} $$
i.e. $x=0, \sqrt{3} y=x+3$ and $\sqrt{3} y=-x-3$
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