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