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

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

But the right answer is

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

Where is my error?

Re: Find an equation of the line tangent

If I'm not mistaken your answer is correct.

Re: Find an equation of the line tangent

I agree with Cesc1 - your answer is correct.

- Hollywood

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

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

But the right answer is

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

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:

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

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

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

Contradiction!

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

Re: Find an equation of the line tangent

ah thanks for your explanation

maybe this time the answer paper is wrong xD