Question

At what points do we have a horizontal tangent to the curve `2y^2+4x^2-y=16x`?

Answer 1

Tangent is horizontal at `(2,3.09)` and `(2,-2.59)`

Explanation:

The function `2y^2+4x^2-y=16x` will have a horizontal tangent line, where slope of the curve i.e. derivative of function equals zero.

Hence let us first find the derivative of `2y^2+4x^2-y=16x`. Differentiating (considering it an implicit function), we get

`4y(dy)/(dx)+8x-(dy)/(dx)=16`

or `(dy)/(dx)(4y-1)=16-8x`

or `(dy)/(dx)=(16-8x)/(4y-1)=(8(2-x))/(4y-1)`

and this is zero when `x=2`.

When `x=2`, we have `2y^2+4*2^2-y=16xx2`

or `2y^2-y+16-32=0`

or `2y^2-y-16=0`

i.e. `y=(1+-sqrt(1+128))/4=(1+-sqrt129)/4`

i.e. `3.09` or `-2.59`

Hence tangent is horizontal at `(2,3.09)` and `(2,-2.59)`

graph{(2y^2+4x^2-y-16x)((x-2)^2+(y-3.09)^2-0.02)((x-2)^2+(y+2.59)^2-0.02)=0 [-10, 10, -5, 5]}