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  • Let a common tangent to the curves    and <img alt="(x-4)^{2}+y^{2} = 16" src="https://

Let a common tangent to the curves  y^{2}2=4x  and (x-4)^{2}+y^{2} = 16  touch the curves at the points P and Q.
Then  (PQ)^{2}  is equal to ____ : 

Option: 1

32


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

y^{2}=4x\\

(x-4)^{2}+y^{2}=16

Let equation of tangent of parabola
y=mx+1/m      ....(1)
Now equation 1 also touches the circle

\begin{aligned} & \therefore\left|\frac{4 m+1 / m}{\sqrt{1+m^2}}\right|=4 \\ & (4 \mathrm{~m}+1 / \mathrm{m})^2=16+16 \mathrm{~m}^2 \\ & 16 \mathrm{~m}^4+8 \mathrm{~m}^2+1=16 \mathrm{~m}^2+16 \mathrm{~m}^4 \\ & 8 \mathrm{~m}^2=1 \\ & \mathrm{~m}^2=1 / 8 \quad\left\{m^4=0\right\}(\mathrm{m} \rightarrow \infty) \end{aligned}

For distinet points consider only  m^{2}=1/8. 

Point of contact of parabola

\begin{aligned} & P(8,4 \sqrt{2}) \\ & \therefore P Q=\sqrt{S_1} \Rightarrow(P Q)^2=S_1 \\ & =16+32-16=32 \end{aligned}

 

Posted by

Gaurav

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