The equation `5x^2 - 6xy + 5y^2 = 16`  represents an ellipse. Determine two points on the ellipse at which the tangent is horizontal.

1 Answer | Add Yours

Top Answer

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

`5x^2 - 6xy + 5y^2 = 16`

`` To find the points we need to find the derivative by using implicit differentiation and then find the derivatives zero.

==> `10x - 6y - 6xy' + 10yy' `

`==gt 10yy'- 6xy' = 6y-10x `

`==gt y' (10y - 6x) = 6y-10x `

``

`=gt 6y - 10x = 0 `

`==gt y= 10x/6 = (5x)/3`

`` ==> Now we will substitute `y= (5x)/3`  into the equation.

==> `5x^2 - 6xy + 5y^2 = 16 `

`==gt 5x^2 - 6x(5x/3) + 5(5x/3)^2 = 16 `

`==gt 5x^2 - 10x^2 + 125x^2 /9 = 16`

`` ==>  Multiply by 9.

==> `45x^2 - 90x^2 + 125x^2 = 144 `

`==gt 80x^2= 144 `

`==gt x^2 = 144/80 = 9/5 `

`==gt x = +- 3/sqrt5 = 3sqrt5/5 `

`==gt y= 5x/3 ==gt y= +-sqrt5`

`` Then the points such that the tangent is horizontal are:

`((3sqrt5)/5 , sqrt5 ) and ((-3sqrt5)/5 , -sqrt5)`

We’ve answered 318,928 questions. We can answer yours, too.

Ask a question