tangent line prob: A rotated hyperbola is defined implicitly

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xomandi

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A rotated hyperbola is defined implicitly by the equation 2x^2 + 9xy + 4y^2 - 8x = 36.

a) Write the equation of the tangent to this curve at the point (0,3).

b) How many other tangent lines are also parallel to the tangent line you found in part (a)?

The derivative I got for this is:

4x + 9y + y' * 9x + 8y - 8 = 0

Then I should solve for y', right? And y' should be the slope of the tangent line?

This really meant to be a question but thinking about it more I think this is the right way.

Thanks for help in advance!
 
Two Things:

1) (0,3) is not on the curve. Perhaps it is worded badly and this is not the intent?

2) Your implicit differentiation has gone astray. "8y"?

These objections were outdated when the correct equation was posted. See below.
 
\(\displaystyle \L\\2x^{2}+9xy+4y^{2}-8x=36\)

Differentiate:

\(\displaystyle \L\\4x+9x\frac{dy}{dx}+9y+8y\frac{dy}{dx}-8=0\)

Solve for dy/dx:

\(\displaystyle \L\\\frac{dy}{dx}=\frac{8-4x-9y}{9x+8y}\)

Maybe I am erring, but it looks like (0,3) is on the curve:

\(\displaystyle \L\\2(0)^{2}+9(0)(3)+4(3)^{2}-8(0)=36\)

Now, can you find your slope and tangent line at (0,3)?.
 
Ah, they joys of entering the problem correctly. The \(\displaystyle y^{2}\) makes ALL the difference.
 
galactus said:
\(\displaystyle \L\\2x^{2}+9xy+4y^{2}-8x=36\)

Differentiate:

\(\displaystyle \L\\4x+9x\frac{dy}{dx}+9y+8y\frac{dy}{dx}-8=0\)

Solve for dy/dx:

\(\displaystyle \L\\\frac{dy}{dx}=\frac{8-4x-9y}{9x+8y}\)

Maybe I am erring, but it looks like (0,3) is on the curve:

\(\displaystyle \L\\2(0)^{2}+9(0)(3)+4(3)^{2}-8(0)=36\)

Now, can you find your slope and tangent line at (0,3)?.
Click to expand...

yes, i can. 3 = 36(0) + b
b = 3

so its y = 36x + 3

now how about part b?
 
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