Identify the conic section that the given equation represents. x^2 − 3xy + 5y^2 − 28x + 16y − 30 = 0

Answer 1

For given equation find x and y intercepts.

x-intercept substitute y=0 we get x^2 -28x - 30 = 0 this is quadratic, solve for x using quadratic formula

x values are = -b/2a ± (b^2-4ac)^(1/2)/2a, here a = 1, b= -28 and c = -30

= 28/2 ± (784+120)^(1/2)/2 = 14 ± 15.033

x-intercepts -1.033 and 29.033

y-intercept substitute x = 0 we get 5y^2 +16y - 30 = 0 this is quadratic, solve for y using quadratic formula

y values are = -b/2a ± (b^2-4ac)^(1/2)/2a, here a = 5, b= 16 and c = -30

= 16/10 ± (25+600)^(1/2)/2 = -1.6 ± 2.5

y-intercepts -4.1 and 0.9

Since x and y intercepts are unequal the conic section is not circle. I plotted the graph using equation graphing tool. It is an ellipse. The tool is https://www.mathsisfun.com/data/grapher-equation.html

Please acknowledge this answer and tell me if equation graphing tool is useful.

Shailesh K., Senior Math & Physics Instructor

Answer 2

The positive coefficients of the x^2 and y^2 terms, the determinant of

1 -3/2 = 11/4 is positive (notice that we take 1/2 of the cross-terms)

-3/2 5

and the constant being negative guaranteeing feasible solutions on both the x and y axes tells us that we have an ellipse.