Find an Equation of the tangent line to the parabola at the given point and find the x-intercept of the line...
x^2=2y, (4,8)
I haven't learned derivatives yet. Can you solve it like this:
Find the y-intercept by equating the lenths of the two sides of the isosceles triangle formed by the tangent line and the y-axis. (d1 and d2).
I just don't know how to equate them...
and then for slope you would do the slope formula
I suppose that might work but it seems a bit much.
Find the y-intercept of this guy? $\displaystyle \displaystyle f(x) = \frac{x^2}{2}$ , I would make x=0.
Your orignal post didn't mention finding the y-intercept, are we still on the same problem?
You can think of $\displaystyle \displaystyle f'(x) =m = \frac{y_2-y_1}{x_2-x_1}$ for a choosen very small interval around x=4.
let $\displaystyle y = m(x - 4) + 8$ be the equation of the tangent line thru the point $\displaystyle (4,8)$
equation of the parabola is $\displaystyle y = \dfrac{x^2}{2}$
the tangent line and the parabola have equal y-values ...
$\displaystyle \dfrac{x^2}{2} = m(x - 4) + 8
$
$\displaystyle \dfrac{x^2}{2} = mx - 4m + 8$
$\displaystyle \dfrac{x^2}{2} - mx + 4m - 8 = 0$
since the tangent line touches the curve at a single point, the resulting quadratic has a single solution ... the discriminant must = 0
$\displaystyle (-m)^2 - 4\left(\dfrac{1}{2}\right)(4m-8) = 0$
$\displaystyle m^2 - 8m + 16 = 0$
$\displaystyle (m-4)^2 = 0$
choose the positive slope $\displaystyle m = 4$
... now you have the slope of the tangent line and can finish whatever it is you need to do.