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.

How do you find the exact solutions to the system 2x^2+8y^2+8x-48y+30=02x^2+8y^2+8x-48y+30=0 and 2x^2-8y^2=-48y+902x^2-8y^2=-48y+90?