# Thread: Another Tangent line question

1. ## Another Tangent line question

Consider the curve defined by x^2 + 4y^2 = 7 + 3xy

Show that there is a point P with x-coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P

dy/dx = (3y - 2x)/ (8y-3x)

2. Originally Posted by ny_chow
Consider the curve defined by x^2 + 4y^2 = 7 + 3xy

Show that there is a point P with x-coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P

dy/dx = (3y - 2x)/ (8y-3x)
I think you forgot to tell us what the x-coordinate is. For now I'll just call it $\displaystyle x_0$.

So the derivative is $\displaystyle \frac{dy}{dx}=\frac{3y-2x}{8y-3x}$. The horizontal tangent line occurs when the derivative is zero, namely, when $\displaystyle 3y-2x=0$. Since we are given the x-coordinate (which I'm calling $\displaystyle x_0$), we have that $\displaystyle 3y-2x_0=0$. Solving for $\displaystyle y$ gives us $\displaystyle y=\frac{2x_0}{3}$ so the point is $\displaystyle P=\left(x_0,\frac{2x_0}{3}\right)$.

So whatever the x-coordinate actually is, just plug it in for $\displaystyle x_0$ and that will give you the answer.

3. oh right the x coordinate is 3.

yeah thats what i did, but the answer key says to reject y = 2,
and that the answer is 0.165

4. Are you sure you're looking at the right answer? Because y=2 is the only horizontal tangent line to f(x) at x=3. I graphed it to be sure. (It's an ellipse, by the way)

5. To expand a bit on my above answer, it's an ellipse rotated $\displaystyle 22.5^o$ above the x-axis.