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

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March 22nd 2011, 12:03 PM
thelensboss

Find 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)
March 22nd 2011, 12:12 PM
pickslides

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
March 22nd 2011, 12:16 PM
thelensboss

What is f' ?
March 22nd 2011, 12:20 PM
pickslides

Quote:

Originally Posted by thelensboss

What is f' ?

The derivative of $f(x)$ you can find it like this $f(x)=x^n \implies f'(x)=nx^{n-1}$
March 22nd 2011, 12:27 PM
thelensboss

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
March 22nd 2011, 12:44 PM
pickslides

Quote:

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.
March 22nd 2011, 12:53 PM
thelensboss

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?
March 22nd 2011, 01:06 PM
thelensboss

I need to solve this without CALC as I have not learned derivatives yet.
March 22nd 2011, 04:57 PM
skeeter

Quote:

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.