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)

- Mar 22nd 2011, 12:03 PMthelensbossFind an Equation of the Tangent Line to the parabola at the given point and find....
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) - Mar 22nd 2011, 12:12 PMpickslides
For $\displaystyle \displaystyle f(x)=\frac{x^2}{2}$

The tangent line is $\displaystyle \displaystyle y-f(4) = f'(4)(x-4)$

The x-intercept can be found by making y=0 hence $\displaystyle \displaystyle 0-f(4) = f'(4)(x-4)$ now solve for x - Mar 22nd 2011, 12:16 PMthelensboss
What is f' ?

- Mar 22nd 2011, 12:20 PMpickslides
- Mar 22nd 2011, 12:27 PMthelensboss
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 - Mar 22nd 2011, 12:44 PMpickslides
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. - Mar 22nd 2011, 12:53 PMthelensboss
Alright thanks! But I still don't know how to find m with this? I just choose something? I guess I would use (4,8) and another point but what do I use?

- Mar 22nd 2011, 01:06 PMthelensboss
I need to solve this without CALC as I have not learned derivatives yet.

- Mar 22nd 2011, 04:57 PMskeeter
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.