Question

What is the equation of the tangent and point of contact between `y^2=4ax` and a line parallel to `2x+y=0`?

Answer 1

tangent equation: `x=-2(y+2a)`
point of contact: `(4a,-4a)`

Explanation:

In order to use standard methods, this question is easier if we exchange `x` and `y` designations.
To (hopefully) avoid confusion later I will use
`color(white)("XXX")color(blue)(y)=color(red)(barx)`
`color(white)("XXX")color(blue)(x)=color(red)(bary)`

So the parabolic equation becomes:
`color(white)("XXX")barx^2=4abary`
or
`color(white)("XXX")bary= 1/(4a)barx^2`

and the linear equation becomes
`color(white)("XXX")2bary+barx=0`
with a slope `m=(Deltabary)/(Deltabarx)= -2`

note that all linear equations parallel to `2bary+barx=0` must have this same slope, `(-2)`

The general slope of the parabolic equation is given by its first derivative:
`color(white)("XXX")(dbary)/(dbarx) = 1/(2a)barx`

So we have
`color(white)("XXX")1/(2a)barx = -2`

`color(white)("XXX")barx=-4a`

Plugging this into our parabolic equation we get
`color(white)("XXX")bary= (1/(4a))*(-4a)^2 = 4a`

`color(red)((bary,barx)=(4a,-4a))`

To get the equation of the tangent line, note
`color(white)("XXX")(y-bary)/(x-barx) = -2`

`color(white)("XXX")bary-4a=(-2)(barx+4a)`

`color(white)("XXX")bary = -2barx-8a + 4a = (-2)(barx+2a)`

`color(red)(bary = (-2)(barx+2a))`

Reverting to the original variables:
`color(white)("XXX")(color(blue)(x),color(blue)(y)) = (4a,-4a)`
and
`color(white)("XXX")color(blue)(x)=(-2)(color(blue)(y)+2a)`