Question

23. Find the angle between the curves 2y^2 – 9x = 0, 3x^2 + 4y = 0(in the 4th quadrant).​

Answer 1

The angle between the curves 2y^2 – 9x = 0, 3x^2 + 4y = 0 is $tan^{-1}$(9/13)

First lets find the point of intersection of the two curves.

9x = 2y^2 == >

x = 2y^2/9

Substitute x in second curve:

3 ( 2y^2/9)^2 + 4y = 0

==> y( 12y^3/81 + 4) = 0

• ==> y = 0 and y^3 = 81 x -4 /12 = -27

• ==> y = 0 and y = -3

x = 2y^2/9

• y = 0 ==> x = 0 == > ( 0 , 0 )

• y = -3 ==> x = 2 == > ( 2 , -3 ) , 4th quadrant

Lets find the slopes of the 2 curves.

m1 = dy/dx

dx/dy = 4y/9 ==> dy/dx = 9/4y = -3/4 at ( 2 , -3 )

m2 = dy/dx

dy/dx = - 6x/4 =  -3

Therefore the angle between the two curves is

• tan $\theta$ = $|\frac{m1 - m2}{1 + m1m2} |$ = $|\frac{ -3/4 - -3}{ 1 + -3/4*-3} |$ = $|\frac{9/4}{1 + 9/4} | = |\frac{9/4}{13/4} | = |9/13|$

Angle between the two curves is  $tan^{-1}$(9/13)