Conic Sections Question 117
Question: The locus of the point of intersection of the perpendicular tangents to the ellipse $ \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 $ is
[Karnataka CET 2003]
Options:
A) $ x^{2}+y^{2}=9 $
B) $ x^{2}+y^{2}=4 $
C) $ x^{2}+y^{2}=13 $
D) $ x^{2}+y^{2}=5 $
Show Answer
Answer:
Correct Answer: C
Solution:
The locus of point of intersection of two perpendicular tangents drawn on the ellipse is $ x^{2}+y^{2}=a^{2}+b^{2}, $ which is called -director- circle…… Given ellipse is $ \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 $ , \Locus is $ x^{2}+y^{2}=13. $
BETA
