Chapter SH, Problem 9.15EP

To determine

To find: whether the equation is parabola, circle, ellipse or hyperbola without writing in the standard form.

Answer to Problem 9.15EP

The equation is parabola.

Explanation of Solution

Given:

  ${\text{8y}}^2\text{-16x+12y-20=0}$ .

Concept used:

The equation of conic is in the form:

  ${\text{Ax}}^{\text{2}}{\text{+Bxy+Cy}}^{\text{2}}\text{+Dx+Ey+F=0}$ .

conic discriminant= ${\text{B}}^{\text{2}}\text{-4AC}$ .

If ${\text{B}}^{\text{2}}\text{-4AC<0}$ then it’s ellipse.

If ${\text{B}}^{\text{2}}\text{-4AC=0}$ then it’s parabola.

If ${\text{B}}^{\text{2}}\text{-4AC>0}$ then its hyperbola.

Calculation:

The equation is here

  ${\text{8y}}^2\text{-16x+12y-20=0}$ .

The equation of conic is in the form:

  ${\text{Ax}}^{\text{2}}{\text{+Bxy+Cy}}^{\text{2}}\text{+Dx+Ey+F=0}$ .

Where,

  $\text{A}$ is the coefficient of the ${\text{x}}^2$ .

  $\text{B}$ is the coefficient of the $\text{xy}$ .

  $\text{C}$ is the coefficient of the ${\text{y}}^2$ .

Here $\text{A=0,B=0 and C=8}$ .

According to the formula:

  ${\text{B}}^{\text{2}}\text{-4AC=0-4}\left(0\right)\left(8\right)=0.$

If ${\text{B}}^{\text{2}}\text{-4AC=0}$ then it’s parabola.

Hence, the equation is parabola.