Conic Sections Question 95
Question: The equation of common tangent to the circle $ x^{2}+y^{2}=2 $ and parabola $ y^{2}=8x $ is
[RPET 1997]
Options:
A) $ y=x+1 $
B) $ y=x+2 $
C) $ y=x-2 $
D) $ y=-x+2 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y^{2}=8x, $
$ \therefore 4a=8 $
therefore $ a=2 $
Any tangent of parabola is, $ y=mx+\frac{a}{m} $ or $ mx-y+\frac{2}{m}=0 $
If it is a tangent to the circle $ x^{2}+y^{2}=2, $ then perpendicular from centre $ (0,0) $ is equal to radius $ \sqrt{2} $ .
$ \therefore \frac{2/m}{\sqrt{m^{2}+1}}=\sqrt{2} $ or $ \frac{4}{m^{2}}=2(m^{2}+1) $
therefore $ m^{4}+m^{2}-2=0 $
therefore $ (m^{2}+2)(m^{2}-1)=0 $ or $ m=\pm 1 $
Hence the common tangent are $ y=\pm (x+2) $
$ \therefore y=x+2 $ .
BETA
