# "hard" implict tangent question

• Oct 17th 2007, 03:57 PM
polymerase
"hard" implict tangent question
Find all the ponts on the curve $(x^2 + y^2)^2=8(x^2-y^2)$ where the tangent line is horizontal. I've graphed it, i know the answer is $(-1.75,1),(1.75,1),(-1.75,-1),(1.75,-1)$ but how do u get that mathematically?
• Oct 17th 2007, 04:28 PM
topsquark
Quote:

Originally Posted by polymerase
Find all the ponts on the curve $(x^2 + y^2)^2=8(x^2-y^2)$ where the tangent line is horizontal. I've graphed it, i know the answer is $(-1.75,1),(1.75,1),(-1.75,-1),(1.75,-1)$ but how do u get that mathematically?

Your points are close, but not quite.

Take the derivative:
$2(x^2 + y^2) \left ( 2x + 2y\frac{dy}{dx} \right ) = 8 \left ( 2x - 2y \frac{dy}{dx} \right )$

The slope of the tangent is horizontal when $\frac{dy}{dx} = 0$

Sooooo....
$2(x^2 + y^2) ( 2x + 2y\cdot 0 ) = 8 ( 2x - 2y \cdot 0 )$

$2(x^2 + y^2) (2x) = 8 (2x)$

$4x^3 + 4xy^2 = 16x$

$x^2 + y^2 = 4$

So the points where we have a 0 slope tangent are the points where the original function intersects the circle $x^2 + y^2 = 4$.

So $y^2 = 4 - x^2$ is our condition we insert into the original relation.

Thus
$(x^2 + (4 - x^2))^2=8(x^2-(4 - x^2))$

You can solve this for the x and y values.

-Dan