Thread: 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)

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 .

So the derivative is . The horizontal tangent line occurs when the derivative is zero, namely, when . Since we are given the x-coordinate (which I'm calling ), we have that . Solving for gives us so the point is .

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

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

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)

To expand a bit on my above answer, it's an ellipse rotated above the x-axis.