The condition that the line x = my + c may be a tangent of x2a2-y2b2=-1 is

A
$c^{2} = a^{2} m^{2} - b^{2}$
B
$c^{2} = a^{2} - b^{2} m^{2}$
C
$c^{2} = b^{2} - a^{2} m^{2}$
D
$c^{2} = b^{2} m^{2} - a^{2}$

Solution:

$G i v e n h y p e r b o l a \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1 Given tangent x = my + c \Rightarrow x - my - c = 0 — ① let P (x_{1}, y_{1}) be point of contact . Now tangent at P is S_{1} = 0 \Rightarrow \frac{x x_{1}}{a^{2}} - \frac{y y_{1}}{b^{2}} + 1 = 0 — ② Since ① & ② represents same line then \Rightarrow \frac{x_{1}}{\frac{a^{2}}{1}} = \frac{- y_{1}}{\frac{b^{2}}{- m}} = \frac{1}{- c} \Rightarrow x_{1} = \frac{- a^{2}}{c}, y_{1} = \frac{- b^{2} m}{c} sub point in line x = my + c \Rightarrow \frac{- a^{2}}{c} = m (\frac{- b^{2} m}{c}) + c \Rightarrow c^{2} = b^{2} m^{2} - a^{2}$

The condition that the line x = my + c may be a tangent of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$ is

Solution:

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