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

2 Answers
Jul 2, 2018

Answer:

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:

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