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]

Question

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]

Smart Achievers · Accepted Answer

(3)

Let the equation of tangent to the parabola $y^{2} = 4 x$ be

$y = m x + \frac{1}{m},$ where $m$ is the slope of the tangent.

If $y = m x + \frac{1}{m}$ is also tangent to the circle ${(x - 3)}^{2} + y^{2} = 9$ ,

then length of perpendicular to the tangent from centre (3, 0) should be equal to radius 3.

i.e., $\frac{3 m + \frac{1}{m}}{\sqrt{m^{2} + 1}} = 3$ $\Rightarrow 9 m^{2} + \frac{1}{m^{2}} + 6 = 9 m^{2} + 9$ $\Rightarrow m = \pm \frac{1}{\sqrt{3}}$

$∴$ $Tangents are x - \sqrt{3} y + 3 = 0 and x + \sqrt{3} y + 3 = 0$

$out of which x - \sqrt{3} y + 3 = 0 meets the parabola at (3, 2 \sqrt{3}), i.e., above x-axis.$

Solution

Q.
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]

1 $\sqrt{3} y = 3 x + 1$

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

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

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

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Solution

Q. The equation of the common tangent touching the circle (x-3)2+y2=9 and the parabola y2=4x above the x-axis is [2001]

1 3y=3x+1

2 3y=-(x+3)

3 3y=x+3