# Thread: area between the parabola

1. ## area between the parabola

find the area between the parabola y^2 = 4ax and the line y= mx......... even find the equation of the common tangent to the parabolas y^2=4ax and x^2=4by........

2. Area between $\displaystyle y^2 = 4ax$ and $\displaystyle y = mx$

Substitute $\displaystyle x = \frac{y}{m}$ so we can find the intersection points which we will use for the limits of integration:

$\displaystyle y^2 = 4a \times \frac{y}{m}$

$\displaystyle y^2 = \frac{4ay}{m}$

$\displaystyle y(y - \frac{4a}{m}) = 0$

$\displaystyle y = 0$ or $\displaystyle y = \frac{4a}{m}$

Area = $\displaystyle \int_0^{\frac{4a}{m}}\, \left( \frac{y}{m} - \frac{y^2}{4a}\right)\, dy$

Can you solve this now?

3. Originally Posted by nzmathman
Area between $\displaystyle y^2 = 4ax$ and $\displaystyle y = mx$

Substitute $\displaystyle x = \frac{y}{m}$ so we can find the intersection points which we will use for the limits of integration:

$\displaystyle y^2 = 4a \times \frac{y}{m}$

$\displaystyle y^2 = \frac{4ay}{m}$

$\displaystyle y(y - \frac{4a}{m}) = 0$

$\displaystyle y = 0$ or $\displaystyle y = \frac{4a}{m}$

Area = $\displaystyle \int_0^{\frac{4a}{m}}\, \left( \frac{y}{m} - \frac{y^2}{4a}\right)\, dy$

Can you solve this now?
Alternatively, in order to get it in familiar terms (i.e. in terms of x), substitute $\displaystyle y = mx$ and find the intersection points.

I.e. $\displaystyle (mx)^2 = 4ax$

$\displaystyle m^2x^2 - 4ax = 0$

$\displaystyle x(m^2 x - 4a) = 0$

$\displaystyle x= 0$ or $\displaystyle x = \frac{4a}{m^2}$.

So Area = $\displaystyle \int_0^{\frac{4a}{m^2}}{(mx - 2\sqrt{a}\sqrt{x})\,dx}$.

4. Yes that's what I originally started out doing, but was trying to avoid integration of the square root.

5. Originally Posted by nzmathman
Yes that's what I originally started out doing, but was trying to avoid integration of the square root.
Why?

Integrating a square root is easy...

Remember that $\displaystyle \sqrt{x} = x^{\frac{1}{2}}$

So $\displaystyle \int{\sqrt{x}\,dx} = \int{x^{\frac{1}{2}}\,dx}$

$\displaystyle = \frac{1}{\frac{1}{2}}x^{\frac{1}{2}-1} + C$

$\displaystyle = 2x^{-\frac{1}{2}} + C$

$\displaystyle = \frac{2}{x^{\frac{1}{2}}} + C$

$\displaystyle = \frac{2}{\sqrt{x}} + C$

$\displaystyle = \frac{2\sqrt{x}}{x} + C$.

You just need to remember your index laws

6. Actually,

$\displaystyle \int \sqrt{x} dx = \frac{2}{3} x^{\frac{3}{2}} + C= \frac{2}{3} x \sqrt{x} + C$

I know it's easy but I was trying to make the integration a little easier for the person who posed the question.

7. Originally Posted by nzmathman
Actually,

$\displaystyle \int \sqrt{x} dx = \frac{2}{3} x^{\frac{3}{2}} + C= \frac{2}{3} x \sqrt{x} + C$

I know it's easy but I was trying to make the integration a little easier for the person who posed the question.
Oh - oops! My bad lol... Yes you are right :P

8. ## Common tangents to parabolas

Hi-

Originally Posted by Shivanand
... even find the equation of the common tangent to the parabolas y^2=4ax and x^2=4by........
Have you not looked at my solution to your earlier posting? http://www.mathhelpforum.com/math-he...-parabola.html

I promise you - it is correct!