Parabola and Line Intersection Points

**Stevo_Evo_22** · March 14th 2009, 10:29 PM

Here is the problem:
A graphic designer draws a logo involving a parabola ‘sitting’ in a V shape on a set of axes as shown at right.

Find the equation of the parabola, given it is of the form y = kx2 and the points of intersection of the V with the parabola.

You probably can't see the picture, so the equations of the lines are -2x-2 and 2x-2, and the only point I have for the parabola is (0,0) as the turning point. I'm having trouble with the first part, finding the equation of the first point, because I can't find k. If I can get this bit, I can work the rest out

Thanks,
Steven

**Grandad** · March 14th 2009, 11:10 PM

Hello Steven

Originally Posted by Stevo_Evo_22

Here is the problem:
A graphic designer draws a logo involving a parabola ‘sitting’ in a V shape on a set of axes as shown at right.

Find the equation of the parabola, given it is of the form y = kx2 and the points of intersection of the V with the parabola.

You probably can't see the picture, so the equations of the lines are -2x-2 and 2x-2, and the only point I have for the parabola is (0,0) as the turning point. I'm having trouble with the first part, finding the equation of the first point, because I can't find k. If I can get this bit, I can work the rest out

Thanks,
Steven

Although you haven't said so, I'm guessing that the lines $y=\pm 2x - 2$ are tangents to the parabola. So, on that assumption, we can find the tangent at the point $(x_1, kx_1^2)$ , and then compare it to one of these equations. Like this:

$y = kx^2$

$\Rightarrow \frac{dy}{dx}= 2kx =2kx_1$ at $(x_1, kx_1^2)$

So the tangent at this point is

$y-kx_1^2 =2kx_1(x-x_1)$

i.e. $y =2kx_1x -kx_1^2$

So if this is the same as the line $y = 2x -2$ , we can compare coefficients to get:

$2kx_1 =2$ and $-kx_1^2 = -2$

$\Rightarrow kx_1= 1$ and $kx_1^2 =2$

$\Rightarrow k =0.5$ and $x_1 = 2$

Can you complete it now?

Grandad

**Stevo_Evo_22** · March 15th 2009, 01:43 AM

Ok now I'm really lost

Sorry I don't know when you're talking about the parabola or the line...

**Grandad** · March 15th 2009, 03:22 AM

Hello Steven

Originally Posted by Grandad

Although you haven't said so, I'm guessing that the lines $y=\pm 2x - 2$ are tangents to the parabola. So, on that assumption, we can find the tangent at the point $(x_1, kx_1^2)$ , and then compare it to one of these equations. Like this:

$y = kx^2$

$\Rightarrow \frac{dy}{dx}= 2kx =2kx_1$ at $(x_1, kx_1^2)$

So the tangent at this point is

$y-kx_1^2 =2kx_1(x-x_1)$

i.e. $y =2kx_1x -kx_1^2$ This is the equation of the tangent to the parabola at any general point $\color{red}(x_1, kx_1^2)$ on the parabola. Do you understand how I got this?

So if this is the same as the line $y = 2x -2$ (i.e. one of the two lines you were given in the question), we can compare coefficients to get:

Coefficients of $\color{red}x$ in the two equations are the same: $2kx_1 =2$ and the constant terms are the same: $-kx_1^2 = -2$

$\Rightarrow kx_1= 1$ and $kx_1^2 =2$

Divide the second equation by the first: $\color{red}\frac{kx_1^2}{kx_1} = \frac{2}{1}$

$\Rightarrow k =0.5$ and $x_1 = 2$

Can you complete it now?

Grandad

I've put further explanations in red. What this shows is that the parabola $y = 0.5x^2$ has a tangent at the point $(2, 2)$ whose equation is $y = 2x -2$ ; i.e. one of the lines that you were given. Because the parabola is symmetrical about the y-axis, the tangent at $(-2,2)$ will be the other line you were given: $y = -2x -2$ .

Is that OK now?

Grandad

**Stevo_Evo_22** · March 16th 2009, 09:19 PM

Thankyou very much for your help, and sorry I couldn't get back to you sooner

Unfortunately I still do not understand how you did it, because I've never seen something like that before, however I was able to work out the question using the discriminant.

Thanks again,
Steven

**Grandad** · March 17th 2009, 01:50 AM

Hello Steven

Sorry if my explanation used stuff you haven't covered in your course. I assumed that you had done some calculus, which it now appears you haven't.

Yes, you can solve the problem without calculus using the discriminant of a quadratic equation. I imagine you said something like this:

The curve $y = kx^2$ meets the line $y = 2x - 2$ where

$kx^2 = 2x - 2$

$\Rightarrow kx^2 - 2x +2 = 0$

The line is a tangent to the curve if the roots of this equation are equal. (This is the hard bit - do you understand it?)

i.e. if $(-2)^2 - 4\cdot k \cdot 2 = 0$

$\Rightarrow 8k = 4$

$\Rightarrow k = 0.5$ , which is the same answer as I got using calculus methods.

$\Rightarrow x = ...$ etc.

Grandad

**Stevo_Evo_22** · March 17th 2009, 12:39 PM

Yes you're correct-I have yet to do calculus

So as you said, the line only touches the curve so the discriminant must equal 0 for there to only be one solution. All I had to do was simply solve for k as you did with calculus...

Thanks!

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