Question

How do you determine circle, parabola, ellipse, or hyperbola from equation `4x^2 + 9y^2 - 16x +18y -11 = 0`?

Answer 1

See below

Explanation:

`4x^2 + 9y^2 - 16x +18y -11 = 0`

Here's an easy way:
-If the coefficients on `x^2` and `y^2` match, it is a circle

-If there is only one squared term, it is a parabola

-If one of the squared terms has a negative coefficient, it is a hyperbola

-If the coefficients on `x^2` and `y^2` don't match but they still have coefficients that either both positive or both negative, it is a ellipse

This is an ellipse, let's put in it's standard form:
`4x^2 -16x+ 9y^2 +18y = 11`

`4(x^2 -4x+4)+ 9(y^2 +2y +1)= 11+16+9`

`4(x-2)^2+ 9(y +1)^2= 36`

`(x-2)^2/9+ (y +1)^2/4= 1`

This is a horizontal ellipse

Answer 2

Ellipse

Explanation:

Compare the given general quadratic equation: `4x^2+9y^2-16x+18y-11=0` with the standard form :`ax^2+2hxy+by^2+2gx+2fy+c=0` we get

`a=4, h=0, b=9, g=-8, f=9, c=-11`

Now, the determinant `\Delta` of quadratic equation is given as follows

`\Delta=abc+2fgh-af^2-bg^2-ch^2`

`\Delta=4\cdot9\cdot (-11)+2\cdot 9(-8)(0)-4(9)^2-9(-8)^2-(-11)(0)^2`

`\Delta=-1296\ne0`

hence, the given quadratic equation: `4x^2+9y^2-16x+18y-11=0` represents a conic section

Now, using second determinant

`h^2-ab=0^2-(4)(9)=-36<0`

Given quadratic equation `4x^2+9y^2-16x+18y-11=0` represents an ellipse which can be re-written as follows

`4x^2+9y^2-16x+18y-11=0`

`4(x^2-4x+4)+9(y^2+2y+1)-25-11=0`

`4(x-2)^2+9(y+1)^2=36`

`\frac{4(x-2)^2}{36}+\frac{9(y+1)^2}{36}=1`

`\frac{(x-2)^2}{3^2}+\frac{(y+1)^2}{2^2}=1`

Above ellipse has center at `(2, -1)` & semi-major & semi-minor axes as `3` & `2` respectively