જો અતિવલય frac{{{x^2}}}{4} - frac{{{y^2}}}{5} | vedclass

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Question

Given $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$

$ \Rightarrow {a^2} = 4,{b^2} = 5$

$e = \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2}}}}  = \

Vedclass · Accepted Answer

$ - \frac{{20}}{9}$

Vedclass · Answer

Given $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$

$ \Rightarrow {a^2} = 4,{b^2} = 5$

$e = \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2}}}}  = \sqrt {\frac{{4 + 5}}{4}}  = \frac{3}{2}$

$L = \left( {2 	imes \frac{3}{2},\frac{5}{2}} ight) = \left( {3,\frac{5}{2}} ight)$

equation of tangent at $\left( {{x_1},{y_1}} ight)$ is 

$\frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} = 1$

Here ${x_1} = 3,\,\,\,{y_1} = \frac{5}{2}$

$ \Rightarrow \frac{{3x}}{4} - \frac{y}{2} = 1 \Rightarrow \frac{x}{{\frac{4}{3}}} + \frac{y}{{ - 2}} = 1$

$x$ -intercept pf the tangent, $OA = \frac{4}{3}\,\,$

$y$ -intercept pf the tangent, $\,OB =  - 2$

$O{A^2} - O{B^2} = \frac{{16}}{9} - 4 =  - \frac{{20}}{9}$

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