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The equation of the common tangent touching the cincle $\mathrm{(x-3)^2+y^3=9}$ and the parabola $\mathrm{y^2=4 x}$ above the x-axis is

Option: 1
$\mathrm{\sqrt{3} y-3 x+1}$

Option: 2
$\mathrm{\sqrt{3} y=-(x+3)}$

Option: 3
$\mathrm{\sqrt{3} y=x+3}$

Option: 4
$\mathrm{\sqrt{3} y=-(3 x+1)}$

Answers (1)

best_answer

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

$\mathrm{ y=m x+1 / m }$

It will touch the circle $\mathrm{ (x-3)^2+y^2=9 }$ if distance from the centre $\mathrm{ (3,0) to (1) }$ equals radius of the circle.

i.e. If $\mathrm{ \frac{\mid-0+m(3)+\frac{1}{m}\mid}{\sqrt(-1)^{2}+m^{2}}=3 }$

$\mathrm{ \Rightarrow\left(3 m+\frac{1}{m}\right)^2=9\left(1+m^2\right) }$

$\mathrm{ \Rightarrow 9 m^2+6+\frac{1}{m^2}=9+9 m^2 }$

$\mathrm{ \Rightarrow \frac{1}{m^2}=3 \Rightarrow m= \pm \frac{1}{\sqrt{3}} }$

As the tangent is above x-axis.
We take $\mathrm{ m=\frac{1}{\sqrt{3}} }$ . Hence required tangent is

$\mathrm{ y=\frac{1}{\sqrt{3}} x+\sqrt{3} \text { or } \sqrt{3} y=x+3 . }$

Hence option 3 is correct.

Posted by

HARSH KANKARIA

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