Conic Sections Question 174
Question: For the above problem, the area of triangle formed by chord of contact and the tangents is given by
[Roorkee 1994]
Options:
8
B) $ 8\sqrt{3} $
C) $ 8\sqrt{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Solving above equation with parabola $ y^{2}=4x, $ we get the points $ P(3+2\sqrt{2},2+2\sqrt{2}),Q(3-2\sqrt{2},2-2\sqrt{2}) $ . $ PQ^{2}=32+32=64 $
therefore $ PQ=8 $
Also, if p be perpendicular from (-1,2) on PQ, then area of triangle is $ \frac{1}{2}PQ.p=\frac{1}{2}.8.( \frac{4}{\sqrt{2}} )=8\sqrt{2} $ .
BETA
