” A Java Runtime Environment (JRE) or Java Development Kit (JDK) must be available in order to run Eclipse. No Java virtual machine was found after a search following locations:… Javaw.exe in your current PATH “.

So to resolve ” no java virtual machine was found eclipse … javaw.exe in your current path ” , do the following steps :

1) Download Java Development Kit first from here .

2) Locate javaw.exe file at ” C:\Program Files\Java\jdk-12\bin ”

3) Copy the path and use the short command “Window+ Pause” to open System Information or “Control Panel\System and Security\System”

4) Click on Advanced System settings and open Environment Variables

5) Under System Variable add a new path where you have to put the variable name as PATH and paste the path which you previously copied to Path value

6) Apply changes and now try to run the exe file of Eclipse.

Hope, above instructions helped you.

Solution:

B(t) = ∑^k_i=0   P_k B_k,n(t)         where, 0<=t<=1

It is given that curve passes (0,0) and (-4,2), means starting point of the curve is P₀ = (0,0) and P₃ = (-4,2)

Whereas the curve is controlled by P₁ (14,10) and P₂ (4,0)

Number of control points,

P₀=(0,0) ,  P₁=(14,10) , P₂=(4,0)  , P₃=(-4,2)

Degree of equation n =Control_point -1 = 4 -1 = 3

B(t) = P₀B_0,3(t)+ P₁B_1,3(t) + P₂B_2,3(t)+ P₃B_3,3(t)

Where,  B_k,n(t)=  ⁿC_k    u^k (1-t)^n-k

B_0,3(t)=  ³C₀    t⁰ (1-t)^3-0

=   1(1-t) ³

               =   1(1-t) ³

             =  (1-t)³

B_1,3(t)=  ³C₁    u¹ (1-t)^3-1

=   t(1-t) ²

               =   t(1-t) ²

             =  3t (1-t)²

B_2,3(t)=  ³C₂    t² (1-t)^3-2

=   t²(1-t) ¹

               =   t²(1-t) ¹

             =  3t² (1-t)¹

B_3,3(t)=  ³C₃   u³ (1-t)^3-3

=   t³(1-t) ⁰

               =   t³(1-t) ⁰

             =  t³

B(t)=      P₀ (1-t)³ + P₁ 3t (1-t)²+ P₂ 3t² (1-t)¹+ P₃ t³

Now, Using this equation lets calculate equation for x control point

P₀=(0,0) ,  P₁=(14,10) , P₂=(4,0)  , P₃=(-4,2)

= 0(1-t)³ + (14)  3t (1-t)²+ (4) 3t² (1-t)+ (-4) t³

⁼ 42t(t²+1-2t)+ 12t²(1-t)-4t³

= 42t³+42t-84t²+12t²-12t³-4t³

=26t³-72t²+42t

Using this equation lets calculate equation for y control points

= 0(1-t)³ + (10)  3t (1-t)²+ (0) 3t² (1-t)+ (2) t³

=30t (1+t²-2t)+2t³

=30t+30t³-60t²+2t³

= 32t³ -60t²+30t

So, The answer is P(t) =[ x(t) y(t)] =[ (26t³-72t²+42t) (32t³ -60t²+30t)]