Conic Sections Question 227
Question: The value of m, for which the line $ y=mx+\frac{25\sqrt{3}}{3} $ is a normal to the conic $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1, $ is
Options:
A) $ -\frac{2}{\sqrt{3}} $
B) $ \sqrt{3} $
C) $ -\frac{\sqrt{3}}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The equation of normal to the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ in terms of slope -m- is $ y=mx\pm \frac{m(a^{2}+b^{2})}{\sqrt{a^{2}-b^{2}m^{2}}}; $ Given line $ y=mx+\frac{25\sqrt{3}}{3} $ and coin $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $ which is hyperbola with $ a^{2}=16,b^{2}=9 $ By comparing given line with equation of normal we get $ \pm \frac{m(a^{2}+b^{2})}{\sqrt{a^{2}-b^{2}m^{2}}}=+\frac{25\sqrt{3}}{3} $
$ \Rightarrow \frac{m(16+9)}{\sqrt{16-9m^{2}}}=-\frac{25\sqrt{3}}{3} $
$ \Rightarrow \frac{25m}{\sqrt{16-9m^{2}}}=-\frac{25\sqrt{3}}{3} $
$ \Rightarrow 9m^{2}=3(16-9m^{2})\Rightarrow m^{2}=\frac{16}{12}=\frac{4}{3} $
$ \Rightarrow m=\pm \frac{2}{\sqrt{3}} $
BETA
