Conic-Sections Question 578
Question: The equation of one of the common tangents to the parabola $ y^{2}=8x $ and $ x^{2}+y^{2}-12x+4=0 $ is
Options:
A) $ y=-x+2 $
B) $ y=x-2 $
C) $ y=x+2 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Any tangent to parabola $ y^{2}=8x $ is y $ =mx+\frac{2}{m} $ ? (i) It touches the circle $ x^{2}+y^{2}-12x+4=0 $ , if the length of perpendicular from the centre (6, 0) is equal to radius $ \sqrt{32.} $
$ \therefore \frac{6m+\frac{2}{m}}{\sqrt{m^{2}+1}}=\pm \sqrt{32}\Rightarrow {{( 3m+\frac{1}{m} )}^{2}}=8(m^{2}+1) $
$ \Rightarrow {{(3m^{2}+1)}^{2}}=8(m^{4}+m^{2}) $
$ \Rightarrow m^{4}-2m^{2}+1=0\Rightarrow m=\pm 1 $ Hence, the required tangents are $ y=x+2 $ and $ y=-x-2. $
BETA
