ધારો કે P(6, 3) અતિવલય frac{{{x^2}}}{{{a^2}}} | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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 $(6,3)$

Vedclass · Accepted Answer

$\sqrt {\frac{3}{2}} $

Vedclass · Answer

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 $\frac{9-6}{\frac{6}{a^2}}=\frac{-3}{-\frac{3}{b^2}}$

$\Rightarrow a^2=2 b^2$

$	herefore$ Eccentricity of the hypecbola $=\sqrt{1+\frac{b^2}{a^2}}$

$=\sqrt{1+\frac{b^2}{2 b^2}}$

$=\sqrt{\frac{3}{2}}$

Similar Questions

$\left( {1,\,\,2\,\,\sqrt 2 } \right)$માંથી અતિવલય $16x^{2} - 25y^{2} = 400$ પર દોરેલા સ્પર્શકો વચ્ચેનો ખૂણો.....

રેખા $ ℓx + my + n = 0$ એ અતિવલય $\frac{{{x^2}}}{{{a^2}}}\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1$ નો સ્પર્શક ક્યારે કહેવાય ?

ઉત્કેન્દ્રતા $3/2$ અને નાભિઓ $(\pm 2, 0)$ વાળા અતિવલયનું સમીકરણ :