Find the angle between the parabolas y^2=4ax and x^2=4by at their point of intersection except origin.

Question

Find the angle between the parabolas y^2=4ax  and x^2=4by at their point of intersection except origin.

dls · Answer

$$\LARGE \tan \theta= | \frac{m_1-m_2}{1+m_1m_2}|$$

dls · Answer

$$\LARGE 2y y'=4a =>y' = \frac{2a}{y}$$ that is m1

dls · Answer

$$\LARGE  2x=4by' => y' = \frac{x}{2b}$$
that is m2

dls · Answer

now I have to substitute x and y by solving both the curves together to get the value of m1 m2 right? but im getting weird things while trying to do that..

dls · Answer

If I'm proceeding correctly

dls · Answer

@ganeshie8

ganeshie8 · Answer

looks good, keep going :)

dls · Answer

getting double roots on x..

ganeshie8 · Answer

yes one must be 0 eh ? which you wanto discard anyways..

dls · Answer

yes one is 0 but dunno abt other

ganeshie8 · Answer

y^2 = 4ax -----(1)
x^2 = 4by------(2)

from (2), y = x^2/4b
plug this in (1)

(x^2/4b)^2 = 4ax
solve x

dls · Answer

i did it wait hold on im getting weird stuff :/

ganeshie8 · Answer

(x^2/4b)^2 = 4ax

x^4/16b^2 = 4ax

x(x^3/16b^2-4a) = 0

x= 0, x^3/16b^2-4a = 0

x=0, x^3 = 64ab^2

x=0, x= $4a^{1/3}b^{2/3}$

dls · Answer

that is the weird stuff I meant! :O

dls · Answer

nvm

ganeshie8 · Answer

its not, its just constants a, b..

ganeshie8 · Answer

find y, yes i think this going to turn even more uglier... have fun ! :)