Question

find the angle between the curves 2y^2-9x=0,3x^2+4y=0 (in the 4th quandrant)
plssss ans fast​

Answer 1

The angle between the curves is $tan^{-1}$(1.36)

Step-by-step explanation:

Given as :

The curve equation are

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

Or,  x = $\dfrac{2}{9}$ y²

And

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

Put the value of x from eq 1 into eq 2

i.e  3 ($\dfrac{2}{9}$ y²)² + 4 y = 0

Or, 3 × $\dfrac{4}{81}$ $y^{4}$ + 4 y = 0

Or,  y ( $\dfrac{4}{27}$ y³  + 4 ) = 0

Or,  y = 0   ,  $\dfrac{4}{27}$ y³  + 4 = 0

i.e   $\dfrac{4}{27}$ y³  = - 4

Or, y³ = - 27

∴  y = - 3

Now, put the  value of y into eq 1

x = $\dfrac{2}{9}$ × (-3)²

Or,  x = $\dfrac{2}{9}$ × 9

∴   x = 2

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

Now,

Slope of curve 1 = $m_1$

i.e  $\dfrac{dy}{dx}$  = $m_1$

or,  $\dfrac{d(2 y^{2} - 9 x) }{dx}$ = 2 × 2 $\dfrac{dy}{dx}$  - 9 = 0

Or, 2 × 2 $\dfrac{dy}{dx}$  - 9 = 0

Or, 4 $m_1$ = 9

∴     $m_1$ = $\dfrac{9}{4}$

Slope of first curve = $m_1$ = $\dfrac{9}{4}$

Again

Slope of curve 2  = $m_2$

i.e  $\dfrac{dy}{dx}$  = $m_2$

Or, $\dfrac{d(3x^{2} + 4 y) }{dx}$ = 6 + 4 $\dfrac{dy}{dx}$ = 0

i.e   6 + 4 $\dfrac{dy}{dx}$ = 0

or,  4 $\dfrac{dy}{dx}$ = - 6

Or,  $m_2$ = $\dfrac{-3}{2}$

Slope of second curve = $m_2$ = $\dfrac{-3}{2}$

Now,

Angle between curves

TanФ = $\dfrac{-\dfrac{3}{2} -\dfrac{9}{4} }{1- (\dfrac{-3}{2} )(\dfrac{9}{4}) }$

i.e TanФ = $\dfrac{\dfrac{-13}{4} }{\dfrac{-19}{8} }$

∴ TanФ = $\dfrac{26}{19}$

i.e  Ф  = $tan^{-1}$ ($\dfrac{26}{19}$)

So, The angle between the curves =  Ф  = $tan^{-1}$(1.36)

Hence, The angle between the curves is $tan^{-1}$(1.36)  Answer