Question

If e denotes the eccentricity of the hyperbola, then which of the following statements
is/are TRUE?

Answer 1

ANSWER.

=> option [ A ] and [ D ] is correct answer.

EXPLANATION.

$\sf : \implies \: let \: a \: and \: b \: are \: real \: positive \: number \\ \\ \sf : \implies \: a > 1 \: \: and \: \: \: b < a \: \\ \\ \sf : \implies \: point \: p \: lies \: on \: quadrant \: that \: lies \: on \: hyperbola \: \\ \\ \sf : \implies \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1$

$\sf : \implies \: point \: p \: ( \: a \: \sec( \theta) , \: b \: \tan( \theta) ) \\ \\ \sf : \implies \: equation \: \: are \: = \frac{x x_{1}}{ {a}^{2} } - \: \frac{y y_{1} }{ {b}^{2} } = 1 \\ \\ \sf : \implies \: \frac{x \sec( \theta) }{a} - \frac{y \tan( \theta) }{b} = 1 \\ \\ \sf : \implies \: \frac{ \sec( \theta) }{a} = 1 \\ \\ \sf : \implies \: a \: = \sec( \theta)$

$\sf : \implies \: \dfrac{b \sec( \theta) }{a \tan( \theta) } = 1 \\ \\ \sf : \implies \: b \: = \tan( \theta) \\ \\ \sf : \implies \: eccentricity \: of \: a \: hyperbola \\ \\ \sf : \implies \: {e}^{2} = 1 + \frac{ {b}^{2} }{ {a}^{2} } \\ \\ \sf : \implies \: \: {e}^{2} = 1 + \frac{ \tan {}^{2} ( \theta) }{ \sec {}^{2} ( \theta) }$

$\sf : \implies \: {e}^{2} = 1 + \sin {}^{2} ( \theta) \\ \\ \sf : \implies \: range \: of \: {e}^{2} \: = (1 , 2) \\ \\ \sf : \implies \: 1 < {e}^{2} < 2 \\ \\ \sf : \implies \: 1 < e < \sqrt{2}$

$\sf : \implies \: slope \: of \: normal \: = - 1 \\ \\ \sf : \implies \: equation \: of \: normal \: \implies \: (y - y_{1}) = m(x - x_{1}) \\ \\ \sf : \implies \: (y - \tan {}^{2} \theta) = - 1(x - \sec {}^{2} \theta) \\ \\ \sf : \implies \: y + x = \sec {}^{2} ( \theta) + \tan {}^{2} ( \theta)$

$\sf : \implies \: put \: y \: = 0 \: \: coordinates \: are \\ \\ \sf : \implies \: ( \sec {}^{2} \theta + \tan {}^{2} \theta,0) \\ \\ \sf : \implies \: base \: = \sec {}^{2} \theta + \tan {}^{2} \theta \: - 1 \\ \\ \: \: \sf : \implies \: height \: = \tan {}^{2} ( \theta)$

$\sf : \implies \: area \: of \: \triangle \: = \dfrac{1}{2} \times base \times height \\ \\ \sf : \implies \: \frac{1}{2} \times \tan {}^{2} \theta( \sec {}^{2} \theta + \tan {}^{2} \theta - 1) \\ \\ \sf : \implies \: \frac{1}{2} \times \tan {}^{2} \theta \: ( \tan {}^{2} \theta + \tan {}^{2} \theta) \\ \\ \sf : \implies \: \tan {}^{4} \theta \implies \: \Delta \: = {b}^{4}$