The type of conic sections for the nondegenerate equations given below.a) 6x^{2} + 3x + 10y=10y^{2} + 8b) 3x^{2} + 18xy=5x + 2y + 9c) 4x^{2} + 8x - 5= -y^{2} + 6y + 3

a) Consider the equation ${\displaystyle 6x^2+3x+10y=10y^2+8.}$

This equation can be written as: ${\displaystyle 6x^2-10y^2+10y-8=0}$.

It is known that the equation is in the form of ${\displaystyle A{x}^2+C{y}^2+Dx+Ey+F=0}$

Here, the coefficient of ${\displaystyle x^2}$ and ${\displaystyle y^2}$ are ${\displaystyle A=6}$ and ${\displaystyle C=-10}$, which means ${\displaystyle AC=-60,}$ that is, ${\displaystyle AC<0}$.

Therefore, the equation is of a hyperbola. Hence, the conic section of the given equation is a hyperbola.

b) Consider the equation ${\displaystyle 3x^2+18xy=5x+2y+9}$.

This equation caan be written as: ${\displaystyle 3x^2+18xy-5x-2y-9=0}$, which is in the form of ${\displaystyle A{X}^2+Bxy+C{y}^2+Dx+Ey+F=0}$.

Solve for ${\displaystyle B^2-4AC}$
${\displaystyle B^2-4AC=\left(18\right)-4\left(3\right)\left(0\right)}$
${\displaystyle =324-0}$
${\displaystyle =324}$

This shows that ${\displaystyle B^2-4AC>0.}$ Therefore, the equation is of a hyperbola. Hence, the conic section of the given equation is a heperbola.

c) Consider the equation ${\displaystyle 4x^2+8x-5=-y^2+6y+3}$. T

his equation can written as: ${\displaystyle 4x^2+y^2+8x-6y-8=0}$. It is known that the equation is in the form ${\displaystyle A{X}^2+C{y}^2+Dx+Ey+F=0}$ Here, ${\displaystyle A=4\text{and}C=1,}$ which means ${\displaystyle AC=4\text{and}AC>0.}$

Therefore, the equation is of an ellipse. Hence, the conic section of the given equation is an ellipse.