Parabola Ques 31
- The equation of the common tangent touching the circle $(x-3)^{2}+y^{2}=9$ and the parabola $y^{2}=4 x$ above the $X$-axis is
(2001, 1M)
(a) $\sqrt{3} y=3 x+1$
(b) $\sqrt{3} y=-(x+3)$
(c) $\sqrt{3} y=x+3$
(d) $\sqrt{3} y=-(3 x+1)$
Assertion and Reason
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Solution:
- Any tangent to $y^{2}=4 x$ is of the form $y=m x+\frac{1}{m}$, $(\because a=1)$ and this touches the circle $(x-3)^{2}+y^{2}=9$.
If
$$ \left|\frac{m(3)+\frac{1}{m}-0}{\sqrt{m^{2}+1}}\right|=3 $$
$[\because$ centre of the circle is $(3,0)$ and radius is 3$]$.
$$ \begin{array}{cc} \Rightarrow & \frac{3 m^{2}+1}{m}= \pm 3 \sqrt{m^{2}+1} \\ \Rightarrow & 3 m^{2}+1= \pm 3 m \sqrt{m^{2}+1} \\ \Rightarrow & 9 m^{4}+1+6 m^{2}=9 m^{2}\left(m^{2}+1\right) \\ \Rightarrow & 9 m^{4}+1+6 m^{2}=9 m^{4}+9 m^{2} \\ \Rightarrow & 3 m^{2}=1 \\ \Rightarrow & m= \pm \frac{1}{\sqrt{3}} \end{array} $$
If the tangent touches the parabola and circle above the $X$-axis, then slope $m$ should be positive.
$\therefore m=\frac{1}{\sqrt{3}}$ and the equation is $y=\frac{1}{\sqrt{3}} x+\sqrt{3}$
or
$$ \sqrt{3} y=x+3 \text {. } $$
BETA
