Question

Which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines?

Answer 1

We need to differentiate this with respect to x to see if we can find an expression for the derivative of y at various points.  That will be the slope of the tangent to the curve.  Then we want to see where that derivative might be infinite -- i.e., where the tangent is vertical.

 

It's not written as a function, but it can still be differentiated using the chain rule:

 

x2 + xy + y2 = 3

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

 

(I used parentheses to show the differentiation of each term in the original equation.)

 

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

2x + y = -x dy/dx - 2y dy/dx

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

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

 

We have the derivative of y, but it's defined partly in terms of y itself.  That's OK.  Let's go on...

 

So where would the slope be infinite?  That would happen when x + 2y = 0, or y = -x/2

 

Let's plug that in for y in the original equation to find points where that's the case.

 

x2 + xy + y2 = 3

x2 + x(-x/2) + (-x/2)2 = 3

x2 - x2/2 + x2/4 = 3

3x2 / 4 = 3

x2 = 4

x = ±2

 

So we have two x values where the tangent might be vertical.  Let's plug them into the equation and see what the y values are.  First x = 2...

 

x2 + xy + y2 = 3

4 + 2y + y2 = 3

y2 + 2y + 1 = 0

(y + 1)2 = 0

y = -1

 

So at the point (2, -1) the tangent is vertical.

 

Now try x = -2...

 

x2 + xy + y2 = 3

4 - 2y + y2 = 3

y2 - 2y + 1 =0

(y - 1)2 = 0

y = 1

 

So at the point (-2, 1) the tangent is vertical.