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dolphi86 [110]
3 years ago
12

Find an equation of the tangent line to the hyperbola x2 a2 − y2 b2 = 1 at the point (x0, y0).

Mathematics
1 answer:
nalin [4]3 years ago
3 0

The equation of the tangent line is y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2}  }{a^{2} y_{0} } + y_{0}

To find the tangent to the hyperbola \frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} } at (x₀, y₀), we differentiate the equation implicitly to find the equation of the tangent at (x₀, y₀).

So, \frac{d}{dx} (\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }) = \frac{d0}{dx}\\\frac{d}{dx} \frac{x^{2} }{a^{2} } - \frac{d}{dx}\frac{y^{2} }{b^{2} }= 0\\\frac{2x }{a^{2} } - \frac{dy}{dx}\frac{2y }{b^{2} } = 0\\\frac{2x }{a^{2} } = \frac{dy}{dx}\frac{2y }{b^{2} } \\\frac{dy}{dx} = \frac{b^{2}x }{a^{2}y }

So, at (x₀, y₀)

\frac{dy}{dx} = \frac{b^{2} x_{0} }{a^{2} y_{0} }

So, the equation of the tangent line is gotten from the standard equation of a line in point-slope form

So, \frac{y - y_{0} }{x - x_{0} } = \frac{b^{2} x_{0} }{a^{2} y_{0} }  \\y - y_{0} = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) \\y = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) + y_{0} \\y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2}  }{a^{2} y_{0} } + y_{0}

So, the equation of the tangent line is y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2}  }{a^{2} y_{0} } + y_{0}

Learn more about equation of tangent line here:

brainly.com/question/12561220

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