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10-2. Parabola, Ellipse, Hyperbola
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ધારો કે $P(6, 3)$ અતિવલય $\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1\,$પરનું બિંદુ છે. જો બિંદુ $P$ આગળનો અતિલંબ $x$-અક્ષને $(9, 0),$ આગળ છેદે, તો અતિવલયની ઉત્કેન્દ્રતા :
A
$\sqrt {\frac{5}{2}} $
B
$\sqrt {\frac{3}{2}} $
C
$\sqrt 2 $
D
$\sqrt 3 $
Solution
Equation of normal to the hyperbsla is
$\frac{x-x_1}{\frac{x_1}{a^2}}=\frac{y-y_1}{\frac{-y_1}{b^2}}$
Equation of normal to hyperbola at $(6,3)$ is
$\frac{x-6}{\frac{6}{a^2}}=\frac{y-3}{\frac{-3}{b^2}}$
Since it intersects $x$-axis at $(9,0)$
So $\frac{9-6}{\frac{6}{a^2}}=\frac{-3}{-\frac{3}{b^2}}$
$\Rightarrow a^2=2 b^2$
$\therefore$ Eccentricity of the hypecbola $=\sqrt{1+\frac{b^2}{a^2}}$
$=\sqrt{1+\frac{b^2}{2 b^2}}$
$=\sqrt{\frac{3}{2}}$
Standard 11
Mathematics
