equation of tangent line on curve

**bigwave** · March 17th 2010, 09:23 PM

find an equation of the tangent line to the curve at the given point
(hyperbola)

$\frac{x^2}{16}-\frac{y^2}{9} = 1\ \left(-5,\frac{9}{4}\right)$

using the quotient rule $\frac{vu'-uv'}{v^2}$

$\frac{16(2x)-x^2(0)}{16^2}-\frac{9(2yy')-y^2(0)}{9^2} \Rightarrow \frac{x}{8}-\frac{2yy'}{9} \Rightarrow y'=\frac{9x}{16y} $
so $m=\frac{9(-5)}{16(\frac{9}{4})}\Rightarrow-\frac{5}{4}$

the given answer is $y=-\frac{5}{4}x-4$
but did not see how this was derived (providing my y' was correct)

**drumist** · March 17th 2010, 09:57 PM

I assume you meant to say the given answer is $y=-\frac{5}{4}x-4$

This makes sense with your result, as the slope of the line matches the slope you solved for.

However, you need more information than just the slope. You can use the point-slope form of a line:

$y-y_0 = m (x-x_0)$

In your case, $x_0 = -5$ and $y_0 = \frac{9}{4}$ , so:

$y-\frac{9}{4} = -\frac{5}{4} (x+5)$

Solving for y gives:

$y = -\frac{5}{4}x -\frac{25}{4} + \frac{9}{4}$

$\implies y = -\frac{5}{4}x - \frac{16}{4}$

$\implies y = -\frac{5}{4}x - 4$

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