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 $$\varphi = \,?$$
A.
$${\sin ^{ - 1}}\left( m \right)$$
B.
$${\sin ^{ - 1}}\left( {\frac{a}{{bm}}} \right)$$
C.
$${\sin ^{ - 1}}\left( {\frac{b}{{am}}} \right)$$
D.
$${\sin ^{ - 1}}\left( {\frac{{bm}}{a}} \right)$$
Answer :
$${\sin ^{ - 1}}\left( {\frac{b}{{am}}} \right)$$
Solution :
Equation of tangent at point $$'\varphi '$$ on the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$ is
$$\eqalign{
& \frac{x}{a}\sec \,\varphi - \frac{y}{b}\tan \,\varphi = 1 \cr
& {\text{or }}y = \frac{b}{a}x\,{\text{cosec}}\,\varphi - b\,\cot \,\varphi ......\left( 1 \right) \cr
& {\text{If }}y = mx + \sqrt {{a^2}{m^2} - {b^2}} ......\left( 2 \right) \cr} $$
also touches the hyperbola then on comparing $$\left( 1 \right)$$ & $$\left( 2 \right)$$
$$\eqalign{
& 1 = \frac{{\frac{b}{a}{\text{cosec}}\,\varphi }}{m} = \frac{{ - b\,\cot \,\varphi }}{{\sqrt {{a^2}{m^2} - {b^2}} }} \cr
& {\text{Hence, }}m = \frac{b}{a}{\text{cosec}}\,\varphi \,; \cr
& {\text{or cosec}}\,\varphi = \frac{{am}}{b} \cr
& {\text{or }}\sin \,\varphi = \frac{b}{{am}} \cr
& {\text{or }}\varphi = {\sin ^{ - 1}}\frac{b}{{am}} \cr} $$