Question

Determine the tangent to the Curve 3y^2 = x^3 at (33) and Calculate the area of the triangle bounded by the tangent line the x-axis and the line x=3​

Answer 1

Given : the tangent to the Curve 3y^2 = x^3 at (3,3)

To Find : Tangent Equation

the area of the triangle bounded by the tangent line the x-axis and the line x=3​

Solution:

3y²  = x³

=> 6ydy /dx  = 3x²

=> dy/dx =  x² / 2y

point ( 3 , 3)

Slope  = 3² /2 (2)  = 3/3

Equation of tangent

y  - 3  = (3/2)(x - 3)

=> 2y - 6  = 3x  - 9

=> 2y  = 3x  - 3

2y  = 3x  - 3

x-axis => y = 0  

=> x = 1

( 1 , 0 )    

x = 3   =>  y  = 2

( 3 , 3)

x = 3 and y axis

Hence ( 3 , 0)

So there vertex  of triangle

( 1 , 0) , ( 3 , 0) and ( 3 , 3)

Hence its a right angle triangle

with base = 2 and Height = 3

So area = (1/2) * 2 * 3  =  3 sq units

or using vertex method

= (1/2) | 1 ( 0 - 3) + 3(3 - 0) + 3(0 - 0) |

= (1/2) | - 3 + 9 + 0|

= (1/2)  * 6

= 3 sq units

Hence  area of the triangle bounded by the tangent line the x-axis and the line x=3​    is  3 sq units

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