How do you determine circle, parabola, ellipse, or hyperbola from equation 4x^2 + 9y^2 - 16x +18y -11 = 04x2+9y216x+18y11=0?

2 Answers
Jul 2, 2018

See below

Explanation:

4x^2 + 9y^2 - 16x +18y -11 = 04x2+9y216x+18y11=0

Here's an easy way:
-If the coefficients on x^2x2 and y^2y2 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^2x2 and y^2y2 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 = 114x216x+9y2+18y=11

4(x^2 -4x+4)+ 9(y^2 +2y +1)= 11+16+94(x24x+4)+9(y2+2y+1)=11+16+9

4(x-2)^2+ 9(y +1)^2= 364(x2)2+9(y+1)2=36

(x-2)^2/9+ (y +1)^2/4= 1(x2)29+(y+1)24=1

This is a horizontal ellipse

Ellipse

Explanation:

Compare the given general quadratic equation: 4x^2+9y^2-16x+18y-11=04x2+9y216x+18y11=0 with the standard form :ax^2+2hxy+by^2+2gx+2fy+c=0ax2+2hxy+by2+2gx+2fy+c=0 we get

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