Conic Sections Question 358
Question: The locus of the point of intersection of perpendicular tangents to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ , is
[MP PET 1995]
Options:
A) $ x^{2}+y^{2}=a^{2}-b^{2} $
B) $ x^{2}-y^{2}=a^{2}-b^{2} $
C) $ x^{2}+y^{2}=a^{2}+b^{2} $
D) $ x^{2}-y^{2}=a^{2}+b^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let point be $ (h,k) $ their pair of tangent will be $ ( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1 )( \frac{h^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}-1 )={{( \frac{hx}{a^{2}}+\frac{yk}{b^{2}}-1 )}^{2}} $
Pair of tangents will be perpendicular, if coefficient of $ x^{2} $ + coefficient of $ y^{2}=0 $
therefore $ \frac{k^{2}}{a^{2}b^{2}}+\frac{h^{2}}{a^{2}b^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} $
therefore $ h^{2}+k^{2}=a^{2}+b^{2} $
Replace $ (h,k) $ by $ (x,y) $
therefore $ x^{2}+y^{2}=a^{2}+b^{2} $ .
BETA
