Conic-Sections Question 549
Question: If polar of a circle $ x^{2}+y^{2}=a^{2} $ with respect to $ (x’,y’) $ is $ Ax+By+C=0, $ then its pole will be
Options:
A) $ ( \frac{a^{2}A}{-C},\frac{a^{2}B}{-C} ) $
B) $ ( \frac{a^{2}A}{C},\frac{a^{2}B}{C} ) $
C) $ ( \frac{a^{2}C}{A},\frac{a^{2}C}{B} ) $
D) $ ( \frac{a^{2}C}{-A},\frac{a^{2}C}{-B} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Polar of the circle is $ xx’+yy’=a^{2}, $ but it is given by $ Ax+By+C=0, $ Then $ \frac{x’}{A}=\frac{y’}{B}=\frac{a^{2}}{-C} $ Hence pole is $ ( \frac{a^{2}A}{-C},\frac{a^{2}B}{-C} ) $
BETA
