Question

Find common tangent of the two curves $y^2=4x$ and $x^2+y^2-6x=0$.

• A. $y=\frac{x}{3}+3$
• B. $y=\frac{x}{\sqrt{3}}-\sqrt{3}$
• C. $y=\frac{x}{3}-3$
• D. $y=\frac{x}{\sqrt{3}}+\sqrt{3}$

Answer 1

The correct option is D

$y=\frac{x}{\sqrt{3}}+\sqrt{3}$

Explanation for the correct option:

Step 1. Find the common tangent of two curves $y^2=4x$ and $x^2+y^2-6x=0$

As we know,

$y^2=4x$ is an Equation of parabola

and $x^2+y^2-6x=0$ is an Equation of circle

Step 2. Let $y=mx+\frac{1}{m}$ is the tangent's slope to the parabola that touches the circle $x^2+y^2-6x=0$

Now, centre of the circle $x^2+y^2-6x=0$ is $\left(-g,-f\right)=\left(3,0\right)$

and radius is $r=3$

As we know, the length of tangent from the centre of the circle is equal to the radius of the circle

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

$\Rightarrow$ $9m^4+1+6m^2=9m^2+9m^4$

$\Rightarrow$ $3m^2=1$

$\Rightarrow$ $m=\pm \frac{1}{\sqrt{3}}$

Thus, the required common tangent is $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\pm }\left(\frac{x}{\sqrt{3}}+\sqrt{3}\right)$

Hence, Option ‘D’ is Correct.