Answer:
[tex]x =2[/tex]
[tex]y =2[/tex]
Step-by-step explanation:
Given
[tex]9x^2 + 4(y - 2)^2 = 36[/tex]
[tex]x^2 = 2y[/tex]
Required
Solve
Substitute: [tex]x^2 = 2y[/tex] in [tex]9x^2 + 4(y - 2)^2 = 36[/tex]
[tex]9*2y + 4(y-2)^2 = 36[/tex]
[tex]18y + 4(y-2)^2 = 36[/tex]
Open bracket
[tex]18y + 4[y^2-2y - 2y + 4] = 36[/tex]
[tex]18y + 4[y^2-4y + 4] = 36[/tex]
Open bracket
[tex]18y + 4y^2-16y + 16 = 36[/tex]
Collect like terms
[tex]4y^2+18y-16y + 16 - 36 = 0[/tex]
[tex]4y^2+2y-20= 0[/tex]
Divide through by 2
[tex]2y^2+y-10= 0[/tex]
Expand
[tex]2y^2+5y - 4y-10= 0[/tex]
Factorize
[tex]y(2y + 5) - 2(2y + 5) = 0[/tex]
Factor out 2y + 5
[tex](y -2)(2y + 5) = 0[/tex]
Split
[tex]y - 2 =0\ or\ 2y + 5 = 0[/tex]
Solve for y
[tex]y =2 \ or\y = -\frac{5}{2}[/tex]
We have:
[tex]x^2 = 2y[/tex]
Make x the subject
[tex]x = \sqrt{2y[/tex]
The above is true for positive y values.
So: [tex]y =2[/tex]
This gives
[tex]x = \sqrt{2*2[/tex]
[tex]x = \sqrt{4[/tex]
[tex]x =2[/tex]
