First slide

Hyperbola in conic sections

Question

$If tangent and normal are drawn at the end 'P' of a latus rectum in first quadrant of$
$hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1, meet the transverse axis at T and G respectively, then the area of$ $the triangle P T G is$

Moderate

A

25
B

45
C

90
D

180

Solution

$\begin{array}{l} Given hyperbola is \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1 \\ Here a^{2} = 4, b^{2} = 12 \\ Eccentricity, e = \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} = 2 \\ P = end point of latusrectum (a e, \frac{b^{2}}{a}) \end{array}$
$\begin{array}{l} = (4, 6) \\ Equation of tangent at P is S_{1} = 0 \\ \Rightarrow 2 x - y = 2 i t c u t s x a x i s \Rightarrow y = o \Rightarrow x = 1 \end{array}$
$It meets tranverse axis at T (1, 0)$
$e q u a t i o n o f n o r m a l a t P (4, 6) i s \frac{a^{2} x}{x_{1}} + \frac{b^{2} y}{y_{1}} = a^{2} + b^{2} \Rightarrow \frac{4 x}{4} + \frac{12 y}{6} = 16 \begin{aligned} \end{aligned}$
$\Rightarrow x + 2 y = 16$ $i t c u t s x - a x i s t h e n y = 0, h e n c e x = 16$
$\Rightarrow$ $G = (16, 0)$
$∴ Area of Δ P T G = \frac{1}{2} |\begin{array}{ll} x_{1} - x_{2} & x_{1} - x_{3} \\ y_{1} - y_{2} & y_{1} - y_{3} \end{array}|$

$\begin{aligned} = \frac{1}{2} |\begin{array}{cc} 4 - 1 & 4 - 16 \\ 6 - 0 & 6 - 0 \end{array}| \\ = 45 sq.units \end{aligned}$

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