Question

angles between two curves calculus 2y²=9x and 3x²=-4y​

Answer 1

Answer:

Hi friends

Step-by-step explanation:

2y² - 9x = 0

=> x = 2y²/9 -------(1)

3x² + 4y = 0

=> 3(2y²/9)² + 4y = 0 {x = 2y²/9}

=> 4y⁴/27 + 4y = 0

=> y⁴/27 + y = 0

=> y(y³/27 + 1) = 0

=> y = 0, y = -3

For y = 0 , x = 0

For y = -3 , x = 2

The point (2,-3) lies in the fourth quadrant .

differentiating 2y² - 9x = 0 with respect to x.

=> 4y dy/dx - 9 = 0

=> dy/dx = 9/4y

=> M1 = 9/4×-3 = -3/4

differentiating 3x² + 4y = 0 with respect to x.

=> 6x + 4dy/dx = 0

=> dy/dx = -3x/2

=> M2 = -3

Let the Angle between the curves be x.

tan x = |(m1-m2)/(1+m1×m2)|

=> tan x = 9/13

Angle between curves = x = tan^-1(9/13)

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