Answer: Option B. Ellipse
[tex]\frac{(x-\frac{1}{4})^2}{\frac{129}{16}}+\frac{(y-0)^2}{\frac{129}{32}}=1[/tex]
Step-by-step explanation:
To know what type of conic section the function is
[tex]2x ^ 2-9x + 4y ^ 2 + 8x = 16[/tex] we must simplify it.
[tex]2x ^ 2-9x + 4y ^ 2 + 8x = 16\\\\2x^2 -x +4y^2 =16[/tex]
complete the square of the expression:
[tex]2x ^ 2 -x\\\\\\2(x^2 -\frac{1}{2}x)\\\\2(x^2-\frac{1}{2}x +\frac{1}{16})-2\frac{1}{16}\\\\2(x-\frac{1}{4})^2 -\frac{1}{8}[/tex]
So we have
[tex]2(x-\frac{1}{4})^2 -\frac{1}{8}+4y^2 =16\\\\2(x-\frac{1}{4})^2+4y^2 =\frac{129}{8}\\\\\frac{8}{129}[2(x-\frac{1}{4})^2] +\frac{8}{129}[4y^2] =1\\\\\frac{16(x-\frac{1}{4})^2}{129}+\frac{32(y-0)^2}{129}=1[/tex]
[tex]\frac{(x-\frac{1}{4})^2}{\frac{129}{16}}+\frac{(y-0)^2}{\frac{129}{32}}=1[/tex]
We know that the general equation of an ellipse has the form
[tex]\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2}=1[/tex]
Then the equation
[tex]\frac{(x-\frac{1}{4})^2}{\frac{129}{16}}+\frac{(y-0)^2}{\frac{129}{32}}=1[/tex]
is an ellipse with center [tex](\frac{1}{4}, 0)[/tex]
[tex]a =\sqrt{\frac{129}{16}}[/tex] and [tex]b=\sqrt{\frac{129}{32}}[/tex]
Observe the attached image
