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JEE MainMathematicsModerateConic Sections

Equation of a common tangent to the circle,
x2+y2−6x=0 x^{2} + y^{2} - 6x = 0\ and the parabola, y^2=4 x, is

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Tangent to y^2 = 4x is y = mx +1m\frac{1}{m}
If it is tangent to given circle its distance from (3,0) is equal to 3
 Hence ∣3m+1m1+m2∣=3\ Hence\ \left| \frac{3m + \frac{1}{m}}{\sqrt{1 + m^{2}}} \right| = 3
⇒∣3m2+1∣=3m1+m2\Rightarrow \left| 3m^{2} + 1 \right| = 3m\sqrt{1 + m^{2}}
squaring both sides, we get
9m4+6m2+1=9m2+9m4⇒m=±139m^{4} + 6m^{2} + 1 = 9m^{2} + 9m^{4} \Rightarrow m = \pm \frac{1}{\sqrt{3}}
Hence common tangents are
y=x3+3 or y=−(x3+3)y = \frac{x}{\sqrt{3}} + \sqrt{3}\ or\ y = - \left( \frac{x}{\sqrt{3}} + \sqrt{3} \right)

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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