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જો અતિવલય $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$ ના પ્રથમ ચરણમાં નાભીલંબનો સ્પર્શક $x-$ અક્ષ અને $y-$ અક્ષને અનુક્રમે બિંદુઓ $A$ અને $B$ માં છેદે તો $(OA)^2 - (OB)^2$ = ...................... જ્યાં $O$ એ ઉંગમબિંદુ
$ - \frac{{20}}{9}$
$ \frac{{16}}{9}$
$4$
$ - \frac{{4}}{3}$
Solution
Given $\frac{{{x^2}}}{4} – \frac{{{y^2}}}{5} = 1$
$ \Rightarrow {a^2} = 4,{b^2} = 5$
$e = \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2}}}} = \sqrt {\frac{{4 + 5}}{4}} = \frac{3}{2}$
$L = \left( {2 \times \frac{3}{2},\frac{5}{2}} \right) = \left( {3,\frac{5}{2}} \right)$
equation of tangent at $\left( {{x_1},{y_1}} \right)$ 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, $\,OB = – 2$
$O{A^2} – O{B^2} = \frac{{16}}{9} – 4 = – \frac{{20}}{9}$
