61. Find the equations of the tangent and normal to the hyperbola x2a2−y2b2=1 at the point (x0,y0)

The given eqn of the hyperbola is x2a2−y2b2=1 ______(1) Differentiating eqn (1) wrt 'x' we get, 2xa2−2yb2dydx=0 ⇒2yb2dydx=2xa2 ⇒dydx=b2xa2y dydx|(x,y)=(x0,y0)b2x0a2y0 is the reqd slope of tangent to the curve So, eqn of tangent at point (x0,y0) is y−y0=b2x0a2y0(x−x0). ⇒yy0b2−y02b2=xx0a2−x02a2 ⇒xx0a2−yy0b2=x02a2−y02b2 As (x0,y0) lies on the parabola given by eqn (1) we write, x02a2−y02b2=1 Hence, xx0a2−yy0b2=1

61. 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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Vishal Baghel | Contributor-Level 10

9 months ago

The given eqⁿof the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ ______(1)

Differentiating eqn (1) wrt 'x' we get,

$\frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0$

$\Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}}$

$\Rightarrow \frac{d y}{d x} = \frac{b^{2} x}{a^{2} y}$

${\frac{d y}{d x} |}_{(x, y)} = (x_{0}, y_{0}) \frac{b^{2} x_{0}}{a^{2} y_{0}}$ is the reqd slope of tangent to the curve

So, eqⁿof tangent at point $(x_{0}, y_{0})$ is

$y - y_{0} = \frac{b^{2} x_{0}}{a^{2} y_{0}} (x - x_{0}) .$

$\Rightarrow \frac{y y_{0}}{b^{2}} - \frac{{y_{0}}^{2}}{b^{2}} = \frac{x x_{0}}{a^{2}} - \frac{{x_{0}}^{2}}{a^{2}}$

$\Rightarrow \frac{x x_{0}}{a^{2}} - \frac{y y_{0}}{b^{2}} = \frac{{x_{0}}^{2}}{a^{2}} - \frac{{y_{0}}^{2}}{b^{2}}$

As $(x_{0}, y_{0})$ lies on the parabola given by eqn (1) we write,

$\frac{{x_{0}}^{2}}{a^{2}} - \frac{{y_{0}}^{2}}{b^{2}} = 1$

Hence, $\frac{x x_{0}}{a^{2}} - \frac{y y_{0}}{b^{2}} = 1$

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