Question

How do you find an equation for the tangent line to `x^4=y^2+x^2` at `(2, sqrt12)`?

Answer 1

`y=7/sqrt3x-8/sqrt3` or `sqrt3y=7x-8`

Explanation:

`x^4=y^2+x^2`
`y^2=x^4-x^2`
`2y(dy/dx)=4x^3-2x`
`dy/dx=(4x^3-2x)/(2y)`

at `(2,sqrt12)`
`dy/dx=(4(2^3)-2(2))/(2sqrt12)`
`dy/dx=(32-4)/(4sqrt3)`
`dy/dx=28/(4sqrt3)`
`dy/dx=7/sqrt3`

The equation of tangent, `y-y_1=m(x-x_1)` where `m=7/sqrt3, y_1=sqrt12 and x_1=2`

`y-sqrt12=7/sqrt3(x-2)`
`y-sqrt12=7/sqrt3x-14/sqrt3`
`y=7/sqrt3x-14/sqrt3+sqrt12`
`y=7/sqrt3x-14/sqrt3+sqrt36/sqrt3`
`y=7/sqrt3x-14/sqrt3+6/sqrt3`
`y=7/sqrt3x-8/sqrt3` or
`sqrt3y=7x-8`

Answer 2

`7x-sqrt3y-8=0`. See tangent-inclusive graph. The graph is not to scale. There is contraction in the y-direction.

Explanation:

Differentiating,

`4x^3=2x+2yy'`, giving `y'=7/sqrt3`, at `P(2, sqrt12)`.

The equation to the tangent at P is

y-sqrt13=7/sqrt3(x-2), giving

`7x-sqrt3y-8=0`.

graph{(x^2-sqrt(x^2+y^2))(7x-sqrt3 y-8.2)=0 [-7, 7, -35, 35]}