tangent/ normal lines to curve

[xc630]

Hello, i would appreciate some help with this problem. Thanks in advance.

I need to find the tangent and normal lines to the curve at the given pt.
I know I need to find the slope by finding the derivative

6x^2 + 3xy +2y^2 + 17y-6 =0
Then I got:
12x + 3y + x (dy/dx) + 4 y (dy/dx) + 17 (dy/dx) =0
Then dy/dx = -12x - 3y / 9x+4y+17)

I plugged in the point and got 3/4. However I checked and the answer is 6/7. Can someone show me where I went wrong? thanks

 

[galactus]

Hello, i would appreciate some help with this problem. Thanks in advance.

I need to find the tangent and normal lines to the curve at the given pt.
I know I need to find the slope by finding the derivative

6x^2 + 3xy +2y^2 + 17y-6 =0
Then I got:
12x + 3y + x (dy/dx) + 4 y (dy/dx) + 17 (dy/dx) =0
Then \(\displaystyle \L\\\frac{dy}{dx} = \frac{-12x - 3y}{\underbrace{9}_{\text{3, not 9}}x+4y+17}\)

I plugged in the point and got 3/4. However I checked and the answer is 6/7. Can someone show me where I went wrong? thanks

What's the point?. You forgot to tell us what it is.

 

[xc630]

Sorry the point is (-1, 0) Instead of the nine infornt of the x on teh bottom I meant to type a left parentheses. Why is it 3x on the bottom though? I s it because 3x and the y are the terms multiplied using the product rule?

 

[galactus]

The product rule.

\(\displaystyle \L\\6x^{2}+3xy+2y^{2}+17y-6=0\)

Differentiate implicitly:

\(\displaystyle \L\\12x+\underbrace{3x\frac{dy}{dx}+3y}_{\text{product rule}}+4y\frac{dy}{dx}+17\frac{dy}{dx}=0\)

Solve for dy/dx:

\(\displaystyle \L\\\frac{dy}{dx}=\frac{-3(4x+3y)}{3x+4y+17}\)