How do you find the exact solutions to the system #2x^2+8y^2+8x-48y+30=0# and #2x^2-8y^2=-48y+90#?

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Answer 1 · 2016-12-23T17:02:13

#(x, y)= (-5, 1), (-5, 5)#, and touching double point #(3, 3)#.
Illustrative fine Socratic graph is inserted.

Add and subtract.

#x^2+2x-15=0#, giving #x = -5 and 3#.

#2y^2-12y+15+x=0#.

At #x = -5, y^2-6y+5=0#, giving #y=1 and 5# and,

at #x = 3#, #y^2-6y+9=0#, giving #y=3 and 3#. So,

#(x, y)= (-5, 1), (-5, 5)#, and touching double point #(3, 3)#

graph{(2x^2+8y^2+8x-48y+30)(2x^2-8y^2+48y-90)=0 [-10, 10, -5, 10]}

Note that the intersecting curves are an ellipse and a hyperbola.