Conic Sections Question 211
Question: The line $ y=mx+c $ is a normal to the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}=1 $ , if $ c= $
Options:
A) $ -(2am+bm^{2}) $
B) $ \frac{(a^{2}+b^{2})m}{\sqrt{a^{2}+b^{2}m^{2}}} $
C) $ -\frac{(a^{2}-b^{2})m}{\sqrt{a^{2}+b^{2}m^{2}}} $
D) $ \frac{(a^{2}-b^{2})m}{\sqrt{a^{2}+b^{2}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
As we know that the line $ lx+my+n=0 $ is normal to $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ , if $ \frac{a^{2}}{l^{2}}+\frac{b^{2}}{m^{2}}=\frac{{{(a^{2}-b^{2})}^{2}}}{n^{2}} $ . But in this condition, we have to replace l by m, m by -1 and n by c, then the required condition is $ c=\pm \frac{(a^{2}-b^{2})m}{\sqrt{a^{2}+b^{2}m^{2}}} $ .
BETA
