Gujarati
10-2. Parabola, Ellipse, Hyperbola
easy

The value of m for which $y = mx + 6$ is a tangent to the hyperbola $\frac{{{x^2}}}{{100}} - \frac{{{y^2}}}{{49}} = 1$, is

A

$\sqrt {\frac{{17}}{{20}}} $

B

$\sqrt {\frac{{20}}{{17}}} $

C

$\sqrt {\frac{3}{{20}}} $

D

$\sqrt {\frac{{20}}{3}} $

Solution

(a) If $y = mx + c$ touches $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1,$

then ${c^2} = {a^2}{m^2} – {b^2}$.

Here $c = 6,\,\,{a^2} = 100,\,\,{b^2} = 49$

$\therefore 36 = 100{m^2} – 49$

$\Rightarrow 100{m^2} = 85 $

$\Rightarrow m = \sqrt {\frac{{17}}{{20}}} $.

Standard 11
Mathematics

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