Question

Let `(x_0, y_0)` be a point be a point on the parabola `x^2=4py`. Show that the equation of the line tangent to the parabola at `(x_0, y_0)` is `y = ((x_0/(2p))x) - y_0`?

I need this answer preferably without derivatives.
What I did is y = x^2/4p and y = m(x - x0) + y0 and then solving for m. After solving for m, I plugged it back into y = m(x - x0) = y0 and just ended up with y = x^2/4p. I don't understand what step to take to get to the equation in the problem.

Answer 1

(see below)

Explanation:

`x^2=4pycolor(white)("XXX")rarrcolor(white)("XXX")y=(x^2)/(4p)`
and
for a given point `(x_0,y_0)` on this curve:
[1]`color(white)("XXX")color(blue)(y_0=(x_0)^2/(4p))`

The general slope of the tangent for a point `(x,y)` is given by the derivative:
`color(white)("XXX")m=(dy)/(dx)=2 * x/(4p)=x/(2p)`
and
again, note that for a given point `(x_0,y_0)` on this curve,
[2]`color(white)("XXX")color(magenta)(m_0=(x_0)/(2p))`

For all points, `(x,y)`, on the tangent line, the slope is a constant
`color(white)("XXX")m_0=(y-y_0)/(x-x_0)=color(magenta)((x_0)/(2p))`

This can be re-arranged as:
`color(white)("XXX")y=(x_0(x-x_0))/(2p)+y_0`
or
`color(white)("XXX")y=(x_0)/(2p)xcolor(green)(-((x_0)^2)/(2p))+y_0`

but from [1] we know that `color(blue)((x_0)^2/(4p)=y_0)`
which implies that `color(green)(-((x_0)^2)/(2p)=color(brown)(-2y_0)`

Giving us
`color(white)("XXX")y=x_0/(2p)xcolor(brown)(-2y_0)+y_0`
which simplifies as
`color(white)("XXX")y=(x_0)/(2p)x-y_0`