Question

What are the location of all horizontal and vertical tangents for `x^2-xy^2=49`?

Answer 1

Vertical tangents at the line `x=0`, and the points `(-7,0),(7,0)`; no horizontal tangents

Explanation:

Implicitly differentiate the function. Don't forget to use the product rule for `-xy^2` and that the chain rule will be put into effect when differentiating a function of `y`.

Differentiating gives:

`2x-y^2-x(2y)dy/dx=0`

Solving for `dy/dx` yields:

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

Note that vertical tangent lines will occur whenever `dy/dx` is undefined. In this case, that means the denominator of the function equals `0`.

So, we see that we have vertical tangents when `2xy=0`, meaning `x=0` or `y=0`.

Find the points that correspond with these: when `x=0`, plugging this into the equation, we see that there are no corresponding `y`-values, so there is an asymptote on the line `x=0`.

However, when `y=0`, this yields the equation `x^2=49`, so `x=+-7`, implying vertical tangents at the points `(7,0)` and `(-7,0)`.

To see when there are horizontal tangents, find the times when `dy/dx=0`.

This gives us:

`2x-y^2=0`

Rewrite this using the original function, `x^2-xy^2=49`. When solved for `y^2` this gives `y^2=(x^2-49)/x`. Thus, horizontal tangents occur when:

`2x-(x^2-49)/x=0`

`(2x^2-x^2+49)/x=0`

`x^2+49=0`

This has no real solutions, thus the function never has a horizontal tangent.

Check our answers with a graph:

graph{x^2-xy^2-49=0 [-17.23, 47.7, -14.25, 18.23]}