Parabola 2 Question 5

5. 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: 5. (d)

Solution:

  1. We know that, equation of tangent to parabola $y^{2}=4 a x$ is

$$ y=m x+\frac{a}{m} $$

$\therefore$ Equation of tangent to the parabola $y^{2}=4 x$ is

$$ \begin{aligned} & y=m x+\frac{1}{m} \\ \Rightarrow \quad & m^{2} x-m y+1=0 \end{aligned} $$

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 $=$

$$ (-g,-f) \text { and radius }=\sqrt{\left.g^{2}+f^{2}-c\right]} $$

$\therefore$ The perpendicular distance of $(3,0)$ from the line (i) is 3.

$[\because$ Radius is perpendicular to the

$\Rightarrow \quad \frac{\left|m^{2} \cdot 3-m \cdot 0+1\right|}{\sqrt{\left(m^{2}\right)^{2}+(-m)^{2}}}=3$

tangent of circle]

The length of perpendicular from a point $\left(x _1, y _1\right)$ to the line $a x+b y+c=0$ is $\left|\frac{a x _1+b y _1+c}{\sqrt{a^{2}+b^{2}}}\right|$.

$\Rightarrow \quad \frac{3 m^{2}+1}{\sqrt{m^{4}+m^{2}}}=3$

$\Rightarrow 9 m^{4}+6 m^{2}+1=9\left(m^{4}+m^{2}\right)$

$\Rightarrow \quad m \approx \infty$ 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 $x=0$,

$$ y=\frac{x}{\sqrt{3}}+\sqrt{3} \text { and } y=\frac{-x}{\sqrt{3}}-\sqrt{3} \quad \text { using } y=m x+\frac{1}{m} $$

i.e. $x=0, \sqrt{3} y=x+3$ and $\sqrt{3} y=-x-3$



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