Question

How do you find the coordinates of the points on the curve `x^2-xy+y^2=9` where the tangent line is vertical?

Answer 1

Please see the sketch of a solution below.

Explanation:

Find `dy/dx` using implicit differentiation.

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

The tangent will be vertical when `dy/dx` approaches `oo`, which happens at `y = 1/2x`

Now substitute `1/2x` for `y` in the original equation and solve to get `x=+- 2sqrt3`.

Finish by using `y = 1/2x` to get the `y` coordinates.

So the points are `(2sqrt3, sqrt3)` and `(-2sqrt3, -sqrt3)`

Answer 2

Tangents `x=+-2sqrt3`. Points of contact : `+-sqrt3( 1. 2, )`. See the Socratic tangents-inclusive graph of this ellipse.

Explanation:

x^2+y^2-xy-9=0. represents an ellipse.

In the standard form, this is

`(x+y)^2/36+(x-y)^2/12=1`

Let us find `x'=(dx)/(dy)`.

`2x x'+2y-x'y-x=0,` giving for the vertical direction

`2y=x`, when `x'=1/(y')=1/oo=0`.

Substituting in the equation,

`x^2+x^2/4-x^2/2=9`, giving `x = +-sqrt3`.

The points of contact fo the tangents are `+-sqrt3( 1, 2 )`.

So, the vertical tangents are `x=+-2sqrt3`.

graph{((x+y)^2/36+(x-y)^2/18-1)(x-2sqrt3-.35+.01y)(x+2sqrt3+.35+.01y)=0 [-10, 10, -5, 5]}