# Find an equation of the line tangent

• Mar 9th 2013, 09:40 PM
shiho
Find an equation of the line tangent
Find an equation of the line tangent to the curve $2y^{2}-x^{2}=1$ at the point (7,-5)
I tried to solve it but can't get the right answer, can you help me check it out?

$2y^{2}-x^{2}=1$

Derivative

$4yy' - 2x = 0$
$y'=\frac{x}{2y}$
$y'(7)=\frac{-7}{10}$
$y=mx+b => -5=\frac{-7}{10}7+b=>b=\frac{-1}{10}$

The equation is

$y=\frac{-7}{10}x-\frac{1}{10}$

$y=\frac{-7}{10}x-\frac{17}{2}$

Where is my error?
• Mar 9th 2013, 10:40 PM
Cesc1
Re: Find an equation of the line tangent
• Mar 10th 2013, 01:32 AM
hollywood
Re: Find an equation of the line tangent

- Hollywood
• Mar 10th 2013, 04:31 AM
ILikeSerena
Re: Find an equation of the line tangent
Quote:

Originally Posted by shiho
Find an equation of the line tangent to the curve $2y^{2}-x^{2}=1$ at the point (7,-5)
I tried to solve it but can't get the right answer, can you help me check it out?

$2y^{2}-x^{2}=1$

Derivative

$4yy' - 2x = 0$
$y'=\frac{x}{2y}$
$y'(7)=\frac{-7}{10}$
$y=mx+b => -5=\frac{-7}{10}7+b=>b=\frac{-1}{10}$

The equation is

$y=\frac{-7}{10}x-\frac{1}{10}$

$y=\frac{-7}{10}x-\frac{17}{2}$

Where is my error?

Let's see... suppose the "right answer" is right...
Then the point (7, -5) should be on this tangent line.

So:
$y=\frac{-7}{10}x-\frac{17}{2}$

$-5=\frac{-7}{10}\cdot 7-\frac{17}{2}$

$-5=\frac{-49}{10}-\frac{17}{2} = -13.4$