Find an equation of the line tangent

**shiho** · March 9th 2013, 08:40 PM

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

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

But the right answer is

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

Where is my error?

**Cesc1** · March 9th 2013, 09:40 PM

If I'm not mistaken your answer is correct.

**hollywood** · March 10th 2013, 12:32 AM

I agree with Cesc1 - your answer is correct.

- Hollywood

**ILikeSerena** · March 10th 2013, 03:31 AM

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

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

But the right answer is

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

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:

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

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

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

Contradiction!

So the "right answer" is wrong, or at least it does not belong to the given problem.

**shiho** · March 10th 2013, 08:08 AM

ah thanks for your explanation
maybe this time the answer paper is wrong xD

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