Find the equations of the tangent and normal | vedclass

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(D) Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$,we get $\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0$. Thus,$\

Vedclass · Accepted Answer

Tangent: $\frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$,Normal: $\frac{y-y_{0}}{a^{2} y_{0}}+\frac{x-x_{0}}{b^{2} x_{0}}=0$

Vedclass · Answer

(D) Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$,we get $\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0$. Thus,$\frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$. The slope of the tangent at $(x_{0}, y_{0})$ is $m_{t} = \frac{b^{2}x_{0}}{a^{2}y_{0}}$. The equation of the tangent is $y-y_{0} = \frac{b^{2}x_{0}}{a^{2}y_{0}}(x-x_{0})$,which simplifies to $\frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$. The slope of the normal is $m_{n} = -\frac{1}{m_{t}} = -\frac{a^{2}y_{0}}{b^{2}x_{0}}$. The equation of the normal is $y-y_{0} = -\frac{a^{2}y_{0}}{b^{2}x_{0}}(x-x_{0})$,which can be written as $\frac{y-y_{0}}{a^{2}y_{0}} + \frac{x-x_{0}}{b^{2}x_{0}} = 0$.

Find the equations of the tangent and normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at the point $(x_{0}, y_{0})$.

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