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 2y² - 9x = 0 and 3x² + 4y = 0 is equal to x = tan^-1(9/13).

First we will find the point of contact of the two curves in the 4th quadrant .

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)

Answer 2

The angle between the curve is $Tan^{-1}$ ( $\dfrac{9}{13}$ )

Step-by-step explanation:

Given as :

The two curve equation are

2 y² - 9 x = 0               ........1

3 x²  + 4 y = 0            ...........2

Now, Solving the equation

From eq 2

4 y = - 3 x²

i.e  y = $\dfrac{-3}{4}$ x²

put the value of y in eq 1

2 ($\dfrac{-3x^{2} }{4}$ )² - 9 x = 0

Or, $\dfrac{-9x^{4} }{8}$ = 9 x

Or,  x³ = 8

So, x = 2

For x = 2  , y = $\dfrac{-3}{4}$ (2)²  = - 3

So, The co-ordinate ( x , y ) = ( 2 ,  - 3 )

The co-ordinate is in 4th quadrant

Now, The slope of curves can be determined

From, eq 1

2 y² - 9 x = 0

Slope = $\dfrac{dy}{dx}$ = $m_1$

Differentiate curve with respect to x

2 $\dfrac{dy^{2} }{x}$ - 9 $\dfrac{dx}{dx}$ = 0

Or, 2 $\dfrac{dy^{2} }{dy}$ $\dfrac{dy}{dx}$  -  9 $\dfrac{dx}{dx}$ = 0

Or, 4 y $\dfrac{dy}{dx}$  = 9

At y = - 3

$\dfrac{dy}{dx}$ = $\dfrac{-3}{4}$

i.e  $m_1$  = $\dfrac{-3}{4}$                 ............1

Slope of curve 1 =   $m_1$  = $\dfrac{-3}{4}$

Again

From, eq 2

3 x² + 4 y = 0

Slope = $\dfrac{dy}{dx}$ = $m_2$

Differentiate curve with respect to x

3 $\dfrac{dx^{2} }{x}$ + 4 $\dfrac{dy}{dx}$ = 0

Or, 6 x + 4 $\dfrac{dy}{dx}$ = 0

at x = 2

12 +  4 $\dfrac{dy}{dx}$ = 0

∴  $\dfrac{dy}{dx}$  = $\dfrac{-12}{4}$   =  - 3

i.e $m_2$  = - 3

Slope of curve 2 =  $m_2$  = - 3

So, Angle between the curve = TanФ

TanФ  = $\dfrac{m_1 - m_2}{1+m_1m_2}$

Or, TanФ  = $\dfrac{\dfrac{-3}{4} + 3}{1+(\dfrac{-3}{4} )(-3)}$

Or,   TanФ  = $\dfrac{9}{13}$

∴  Ф  = $Tan^{-1}$ ( $\dfrac{9}{13}$ )

So, The angle between the curve = Ф  = $Tan^{-1}$ ( $\dfrac{9}{13}$ )

Hence, The angle between the curve is $Tan^{-1}$ ( $\dfrac{9}{13}$ )  Answer