Find an Equation of the Tangent Line to the parabola at the given point and find....

**thelensboss** · March 22nd 2011, 12:03 PM

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)

**pickslides** · March 22nd 2011, 12:12 PM

For $\displaystyle f(x)=\frac{x^2}{2}$

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

The x-intercept can be found by making y=0 hence $\displaystyle 0-f(4) = f'(4)(x-4)$ now solve for x

**thelensboss** · March 22nd 2011, 12:16 PM

What is f' ?

**pickslides** · March 22nd 2011, 12:20 PM

Originally Posted by thelensboss

What is f' ?

The derivative of $f(x)$ you can find it like this $f(x)=x^n \implies f'(x)=nx^{n-1}$

**thelensboss** · March 22nd 2011, 12:27 PM

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

**pickslides** · March 22nd 2011, 12:44 PM

Originally Posted by thelensboss

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 suppose that might work but it seems a bit much.

Find the y-intercept of this guy? $\displaystyle f(x) = \frac{x^2}{2}$ , 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 $\displaystyle f'(x) =m = \frac{y_2-y_1}{x_2-x_1}$ for a choosen very small interval around x=4.

**thelensboss** · March 22nd 2011, 12:53 PM

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?

**thelensboss** · March 22nd 2011, 01:06 PM

I need to solve this without CALC as I have not learned derivatives yet.

**skeeter** · March 22nd 2011, 04:57 PM

Originally Posted by thelensboss

I need to solve this without CALC as I have not learned derivatives yet.

let $y = m(x - 4) + 8$ be the equation of the tangent line thru the point $(4,8)$

equation of the parabola is $y = \dfrac{x^2}{2}$

the tangent line and the parabola have equal y-values ...

$\dfrac{x^2}{2} = m(x - 4) + 8<br />$

$\dfrac{x^2}{2} = mx - 4m + 8$

$\dfrac{x^2}{2} - mx + 4m - 8 = 0$

since the tangent line touches the curve at a single point, the resulting quadratic has a single solution ... the discriminant must = 0

$(-m)^2 - 4\left(\dfrac{1}{2}\right)(4m-8) = 0$

$m^2 - 8m + 16 = 0$

$(m-4)^2 = 0$

choose the positive slope $m = 4$

... now you have the slope of the tangent line and can finish whatever it is you need to do.

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