# Modelling the equation of a parabola.

• Jun 19th 2010, 05:41 AM
Shiraz
Modelling the equation of a parabola.
Two parabolas are joined smoothly at the point B(30,20).
a) The parabola through A and B has a zero gradient at A(0,60). Find the equation of this parabola expressing all coefficients as exact values.

The general equation of a quadratic is:
$y = ax^2 + bx + c = 0$
$a(30)^2 + b(30) + c = 20$
$900a + 30b + c = 20$

$a(0)^2 + b(0) + c = 60$
$c = 60$

$\frac{dy}{dx} = 2ax + b = 0$
$= 2a(0) + b = 0$
$b = 0$

$900a + 30b + c = 20$
$900a + 30(0) + 60 = 20$
$900a = -40$
$a = \frac{-2}{45}$

Hence the equation of the parabola is:
$y = \frac{-2}{45}x^2 + 60 = 0$

Is this right so far? Can anyone confirm? Anybody got any ideas on how I'd go about answering b and c?

b) The parabola through B and C has a zero gradient at C. If the equation of this parabola is $y=a(x-p)^2$, find the exact values for a and p.
c) Write down the coordinates of this point.
• Jun 19th 2010, 06:09 AM
Sudharaka
Quote:

Originally Posted by Shiraz
Two parabolas are joined smoothly at the point B(30,20).
a) The parabola through A and B has a zero gradient at A(0,60). Find the equation of this parabola expressing all coefficients as exact values.

The general equation of a quadratic is:
y = ax^2 + bx + c = 0
a(30)^2 + b(30) + c = 20
900a + 30b + c = 20

a(0)^2 + b(0) + c = 60
c = 60

dy/dx = 2ax + b = 0
= 2a(0) + b = 0
b = 0

900a + 30b + c = 20
900a + 30(0) + 60 = 20
900a = -400
a = -2/45

Hence the equation of the parabola is:
y = -2/45x^2 + 60 = 0

Is this right so far? Can anyone confirm? Anybody got any ideas on how I'd go about answering b and c?

b) The parabola through B and C has a zero gradient at C. If the equation of this parabola is y=a(x-p)^2, find the exact values for a and p.
c) Write down the coordinates of this point.

Dear Shiraz,

Correct, except for the fact that you have made a typo in the line I have highlighted above.

For b), since the parabola $y=a(x-p)^2$ goes through B, you can get an equation by substituting this point to the equation. Since $a=-\frac{2}{45}$ you could find p.

Can you continue from here?
• Jun 19th 2010, 08:44 AM
Shiraz
Hi there!
$y= a(x - p)^2$
$y=ax^2 - 2apx + ap^2$
$20 = \frac{-2}{45}(30)^2 - 2(\frac{-2p}{45}(30) + \frac{-2p^2}{45}$
$20 = -40 + \frac{120p}{45} - \frac{2p^2}{45}$
$\frac{-2p^2}{45} + \frac{120p}{45} - 60 = 0$

Though I can't solve for p. Where did I go wrong?
• Jun 19th 2010, 06:18 PM
Sudharaka
Quote:

Originally Posted by Shiraz
Hi there!
$y= a(x - p)^2$
$y=ax^2 - 2apx + ap^2$
$20 = \frac{-2}{45}(30)^2 - 2(\frac{-2p}{45}(30) + \frac{-2p^2}{45}$
$20 = -40 + \frac{120p}{45} - \frac{2p^2}{45}$
$\frac{-2p^2}{45} + \frac{120p}{45} - 60 = 0$

Though I can't solve for p. Where did I go wrong?

Dear Shiraz,

You haven't done anything wrong, but the quadratic equation you obtained does'nt have any real roots. We have taken the value of 'a' (which we obtained before) for the second part. I think this is incorrect since the equation of the first parabola is not given as $y = ax^2 + bx + c$ rather it was taken as such. Hence we cannot use the value of 'a' for part b).

Since $y = a(x-p)^2$ goes through B by substituting the coordinates of B you can get one equation(containing a and p). It had been given that the two parabolas touch smoothly at B. Therefore they must have the same gradient at B. You can obtain the gradient at B from the first parabolic equation. Now differentiate $y = a(x-p)^2$ and you know the value of $\frac{dy}{dx}$ at B. Hope you can continue.
• Jun 19th 2010, 06:31 PM
Shiraz
From (30,20):
$20 = a(30-p)^2$
Also, at (30,20), the gradients of the two parabolas are equal, since the graph joins 'smoothly'.
For $y = -\frac{2}{45}x^2+60$ , $\frac{dy}{dx} = -\frac{4}{45}x$, so at x = 30 the gradient is $-\frac{8}{3}$.
For $y = a(x-p)^2$, $\frac{dy}{dx} = 2a(x-p)$, so at x = 30 the gradient is $2a(30-p)$
So $-\frac{8}{3} = 2a(30-p)$

What do we do next? We still have two unknowns? How can we solve for a and p?
• Jun 19th 2010, 06:50 PM
Sudharaka
Quote:

Originally Posted by Shiraz
From (30,20):
$20 = a(30-p)^2$
Also, at (30,20), the gradients of the two parabolas are equal, since the graph joins 'smoothly'.
For $y = -\frac{2}{45}x^2+60$ , $\frac{dy}{dx} = -\frac{4}{45}x$, so at x = 30 the gradient is $-\frac{8}{3}$.
For $y = a(x-p)^2$, $\frac{dy}{dx} = 2a(x-p)$, so at x = 30 the gradient is $2a(30-p)$
So $-\frac{8}{3} = 2a(30-p)$

What do we do next? We still have two unknowns? How can we solve for a and p?

Dear Shiraz,

Yes. Two unknows and two equations as well. So.......
• Jun 19th 2010, 08:36 PM
Shiraz
Quote:

Originally Posted by Sudharaka
Dear Shiraz,

Yes. Two unknows and two equations as well. So.......

What are these two equations, I'm afraid you've lost me.
• Jun 19th 2010, 10:44 PM
Sudharaka
Quote:

Originally Posted by Shiraz
What are these two equations, I'm afraid you've lost me.

Dear Shiraz,

$20 = a(30-p)^2$---------(1)

$-\frac{8}{3} = 2a(30-p)$--------------(2)