Question

Question #25b60

Answer 1

I found 2 normals:
`y=-sqrt(2)x+4sqrt(2)`
`y=sqrt(2)x-4sqrt(2)`

Explanation:

I would try to derive your function first to find the slope of the curve, `m`:
`2y(dy)/(dx)=4`
`dy/dx=2/y=m` I derived implicitly and found the slope.
Now your function can be written as: `y=+-sqrt(4x)=+-2sqrt(x)`:

So basically at `x=2` we can have 2 slopes:

`m_1=2/y_1=2/(+2sqrt(x))=1/sqrt(x)`
`m_2=2/y_2=2/(-2sqrt(x))=1/-sqrt(x)`

The perpendicular lines in these points will have slopes:

`color(red)(m'=-1/m_(1,2))`

i.e.:
`m'_1=-sqrt(x)`
`m'_2=sqrt(x)`
and equation:
`color(blue)(y-y_0=m'_(1,2)(x-x_0))`

Giving two perpendicular lines in A and B:

In A:
Slope (negative): `m'_1=-sqrt(2)` at `x=2 and y=2sqrt(2)`
Equation:
`y-2sqrt(2)=-sqrt(2)(x-2)`
`y=-sqrt(2)x+2sqrt(2)+2sqrt(2)=-sqrt(2)x+4sqrt(2)`

In B:
Slope (positive): `m'_2=sqrt(2)` at `x=2 and y=-2sqrt(2)`
Equation:
`y+2sqrt(2)=sqrt(2)(x-2)`
`y=sqrt(2)x-2sqrt(2)-2sqrt(2)=sqrt(2)x-4sqrt(2)`