Conic Sections Question 205
Question: If the line $ y=mx+\sqrt{a^{2}m^{2}-b^{2}} $ touches the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ at the point $ \varphi $ . Then $ \varphi $ =
Options:
A) $ {{\sin }^{-1}}(m) $
B) $ {{\sin }^{-1}}( \frac{a}{bm} ) $
C) $ {{\sin }^{-1}}( \frac{b}{am} ) $
D) $ {{\sin }^{-1}}( \frac{bm}{a} ) $
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Answer:
Correct Answer: C
Solution:
[c] Equation of tangent at point $ ‘\varphi ’ $ on the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ is $ \frac{x}{a}\sec \varphi -\frac{y}{b}\tan \varphi =1 $ or $ y=\frac{b}{a}x\cos ec\varphi -b\cot \varphi $
…… (1) If $ y=mx+\sqrt{a^{2}m^{2}-b^{2}} $
…… (2) also touches the hyperbola then on comparing (1) & (2) $ 1=\frac{\frac{b}{a}\cos ec\varphi }{m}=\frac{-b\cot \varphi }{\sqrt{a^{2}m^{2}-b^{2}}} $
Hence, $ m=\frac{b}{a}\cos ec\varphi ; $ or $ \cos ec\varphi =\frac{am}{b} $ Or $ \sin \varphi =\frac{b}{am}, $ or $ \varphi ={{\sin }^{-1}}\frac{b}{am} $
BETA
