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Prove that an equation of the tangent line to the graph of the hyperbola : $(x^2/a^2) - (y^2/b^2) = 1$

at the point ($x_0$, $y_0$) is

$x x_0/a^2 - y y_0/b^2 = 1$ (1)

I implicitly differentiated the equation and then found the gradient by substituting in the points to get the gradient ( $b^2x_0/a^2y_0$) and use the points, plug it into $y-y_1=m(x-x_1)$ But I don't know how to rearrange it to get to (1). Please help me!

sango
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user639649
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    It would be much easier to help if you showed you work – Isaac Browne Apr 07 '19 at 03:22
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    So I differentiated $(x^2/a^2) - (y^2/b^2) = 1$ with respect to x, this was so that I could find the gradient as the question asks to prove that an equation of the tangent line is..... After I implicitly differentiate I get $b^2x/a^2y$. As they have given me the points $(x_0,y_0)$ I sub it into the gradient I have found and that gives me the $(b^2x_0/a^2y_0)$ and now as its asking for equation of a line i sub it into the line formula $y-y_1=m(x-x_1)$, using the gradient and the point I have. I however, am not able to get the equation (1) stated in the question @IsaacBrowne – user639649 Apr 07 '19 at 03:32
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    I saw that. What I meant was showing the process of your substitution. When writing future questions, make sure to include all of your work. This will help your questions get answered better and faster! – Isaac Browne Apr 07 '19 at 03:52

1 Answers1

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\begin{align} y-y_0 & = \frac{b^2x_0}{a^2y_0}(x-x_0) \\ a^2y_0(y-y_0)& = (b^2x_0)(x-x_0) \\ a^2y_0y-a^2y_0^2& = b^2x_0x-b^2x_0^2 \\ b^2x_0^2 -a^2y_0^2& = b^2x_0x-a^2y_0y \end{align} $(x_0,y_0)$ satisfaces that $\frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1$ doing the operations, $b^2x_0^2 -a^2y_0^2=a^2b^2$, then: \begin{align} \\ b^2x_0^2 -a^2y_0^2& = b^2x_0x-a^2y_0y \\ a^2b^2& = b^2x_0x-a^2y_0y \\ 1& = \frac{xx_0}{a^2}-\frac{yy_0}{b^2} \end{align}

sango
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