# Help finding equation of the tangent line to the graph

hauk-gwai

## Homework Statement

Find the equation of the tangent line to the graph of x^2 - xy + y^2 = 19 (where y=y(x)) at (3,-2).

## The Attempt at a Solution

So, this is what I did:
d/dx(x^2-xy+y^2) = (19)d/dx

2x*dx/dx-dy/dx+dy/dx+2y(dy/dx)=0

2x+2y(dy/dx)=0

And I am stuck, the answer to this question is:
y+2=8/7(x-3)

and I don't know how to get the slope.

Homework Helper
Well you have your values of x and y, (3,-2). Sub that into the 2x+2y(dy/dx)=0 you got to solve for dy/dx, then you have the gradient of the tangent. Now use the gradient-point form of a straight line to find the equation of the tangent.

Homework Helper

## Homework Statement

Find the equation of the tangent line to the graph of x^2 - xy + y^2 = 19 (where y=y(x)) at (3,-2).

## The Attempt at a Solution

So, this is what I did:
d/dx(x^2-xy+y^2) = (19)d/dx

2x*dx/dx-dy/dx+dy/dx+2y(dy/dx)=0
d(x^2)/dx= 2x and d(y^2)/dx= 2y(dy/dx) but d(-xy)/dx is NOT -dy/dx+ dy/dx!

2x+2y(dy/dx)=0

And I am stuck, the answer to this question is:
y+2=8/7(x-3)

and I don't know how to get the slope.

hauk-gwai
I am pretty sure I got the x^2 and the y^2 part right.

I am just having trouble taking the deriv of the bolded part:
x^2-xy+y^2=19

Do I apply the product rule for the d/dx(-xy)?

Homework Helper
Yes! It's a product, isn't it?

hauk-gwai
Yeah, but the thing is, I tried applying the product rule, and my results were different, I answer did not match with the answer key, so somewhere within the product rule, I screwed up somewhere, so how do you take d/dx(-xy)?

Staff Emeritus