# Find an equation of the line tangent

• Mar 9th 2013, 08:40 PM
shiho
Find an equation of the line tangent
Find an equation of the line tangent to the curve $\displaystyle 2y^{2}-x^{2}=1$ at the point (7,-5)
I tried to solve it but can't get the right answer, can you help me check it out?

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

Derivative

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

The equation is

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

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

Where is my error?
• Mar 9th 2013, 09:40 PM
Cesc1
Re: Find an equation of the line tangent
• Mar 10th 2013, 12:32 AM
hollywood
Re: Find an equation of the line tangent

- Hollywood
• Mar 10th 2013, 03:31 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 $\displaystyle 2y^{2}-x^{2}=1$ at the point (7,-5)
I tried to solve it but can't get the right answer, can you help me check it out?

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

Derivative

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

The equation is

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

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

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

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

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