Evaluate the ellipse in the equation 4x²+9y²-16x-36y-52=0

Question

Find the elements, graph and label its parts

Answer 1

4x²-16x+9y²-36y+52=0.

Now complete the squares:

4(x²-4x+4)-16+9(y²-4y+4)-36+52=0.

4(x-2)²+9(y-2)²=0, which is not an ellipse. In fact it can only be satisfied when x=y=2, that is, the single point (2,2). To be an ellipse the constant 52 needs to be replaced or removed so that a strictly positive constant can follow the equals sign, and the constant must be less than 52. For example, 48 would create an ellipse, so that we have 4 after the equals. The smaller the constant, the larger the ellipse. The constant 52 causes the ellipse to collapse to a single point.