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:

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3 years ago

You start at (-3,-2). you move down 1 and left 1 unit. where do you end?​

You start at (-3,-2). you move down 1 and left 1 unit. where do you end?