Let a common tangent to the curves y 2 = 4 x and ( x - 4 ) 2 + y 2 = 16 touch the curves at the points P and Q. Then ( P Q ) 2 is equal to _________ . [2023]

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Solution

Q.
Let a common tangent to the curves $y^{2} = 4 x$ and ${(x - 4)}^{2} + y^{2} = 16$ touch the curves at the points P and Q. Then ${(P Q)}^{2}$ is equal to _________ . [2023]

Ans.
(32)

Equation of tangent to parabola $y^{2} = 4 x$ is given by

$y = m x + \frac{1}{m}$ ...(i)

and equation of tangent to the circle ${(x - 4)}^{2} + y^{2} = 16$ is given by

$y = m (x - 4) \pm 4 \sqrt{1 + m^{2}}$ ...(ii)

From (i) and (ii), we get

$\frac{1}{m} = - 4 m \pm 4 \sqrt{1 + m^{2}}$

$\Rightarrow 1 + 4 m^{2} = \pm 4 m \sqrt{1 + m^{2}} \Rightarrow 1 + 16 m^{4} + 8 m^{2} = 16 m^{2} + 16 m^{4}$

$\Rightarrow 8 m^{2} = 1 \Rightarrow m = \pm \frac{1}{2 \sqrt{2}}$

Point of contact of parabola is $(\frac{1}{{(\frac{1}{2 \sqrt{2}})}^{2}}, \frac{2}{\frac{1}{2 \sqrt{2}}})$ $[∵ Point of contact is (\frac{a}{m^{2}}, \frac{2 a}{m})]$

i.e., $(8, 4 \sqrt{2})$

Now, length of tangent $P Q = \sqrt{{(8 - 4)}^{2} + {(4 \sqrt{2})}^{2} - 16}$ $= \sqrt{32}$

$∴$ $P Q^{2} = 32$

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