Help with supposedly simple implicit differentiation

• June 10th 2009, 11:31 AM
Saibot
Help with supposedly simple implicit differentiation
I have a problem with a supposedly easy question...

Find the points on the hyperbola $\frac {x^2} {25} - \frac {y^2} {9} = 1$ where the slope of the tangent is equal to 1.

i end up with $y' = \frac{9x}{25y}$ where y' becomes 1, and i get another function........

the listed answer should be $( \frac{-25}{4} , \frac{-9}{4} ) and ( \frac{25}{4} , \frac{9}{4} )$
• June 10th 2009, 11:46 AM
Amer
Quote:

Originally Posted by Saibot
I have a problem with a supposedly easy question...

Find the points on the hyperbola $\frac {x^2} {25} - \frac {y^2} {9} = 1$ where the slope of the tangent is equal to 1.

i end up with $y' = \frac{9x}{25y}$ where y' becomes 1, and i get another function........

the listed answer should be $( \frac{-25}{4} , \frac{-9}{4} ) and ( \frac{25}{4} , \frac{9}{4} )$

$\frac{x^2}{25} - \frac{y^2}{9} =1$

find y with respect to x

$\frac{x^2}{25} -1 =\frac{y^2}{9}$

$\sqrt{9\left(\frac{x^2}{25} -1\right)} =y$

$y'=\frac{3}{2}\left(\frac{2x}{25}\right)\left(\fra c{1}{(\frac{x^2}{25}-1)^{.5}}\right)$

$y'=\frac{3}{2}\left(\frac{2x}{25\left(\frac{x^2}{2 5}-1\right)^{.5}}\right)=1$

$\frac{3}{2}\left(\frac{2x}{25\left(\frac{x^2}{25}-1\right)^{.5}}\right)=1$

$\frac{3}{2}(2x)=25\sqrt{(\frac{x^2}{25}-1})$

$3(x)=25\sqrt{(\frac{x^2}{25}-1})$

$9x^2=25^2\left(\frac{x^2}{25}-1\right)$

$9x^2=\left(25x^2 - 25^2 \right)$

$25x^2-9x^2=25^2$

$16x^2 = 25^2$ you continue ........

the rest for you
• June 10th 2009, 12:19 PM
Saibot
yea... i tried that method as well... and i get stuck here:
$y' = \frac{2x}{25\sqrt{9(\frac{x^2}{25} - 1)}}$

If we sub y' = 1, how is it possible to solve/isolate x with algebra O_O

edit: nvm, your solution was epic.

nicely done. thanks.
• June 10th 2009, 12:26 PM
Amer
Quote:

Originally Posted by Saibot
yea... i tried that method as well... and i get stuck here:
$y' = \frac{2x}{25\sqrt{9(\frac{x^2}{25} - 1)}}$

If we sub y' = 1, how is it possible to solve/isolate x with algebra O_O

see my old post I edited it best wishes ....

Amer