What is the total number of points of intersection of the graphs of the equations $2 x^{2} - y^{2} = 8, y = x + 2$ $2x^2-y^2=8, y=x+2$ ?

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Answer 1 · 2017-05-03T10:03:22

We have two points $(6, 8)$ $(6,8)$ and $(- 2, 0)$ $(-2,0)$

To find number of points of intersection, we should solve these pair of equations.

As $y = x + 2$ $y=x+2$ , we have $2 x^{2} - {(x + 2)}^{2} = 8$ $2x^2-(x+2)^2=8$

or $2 x^{2} - x^{2} - 4 x - 4 = 8$ $2x^2-x^2-4x-4=8$ or $x^{2} - 4 x - 12 = 0$ $x^2-4x-12=0$

or $x^{2} + 2 x - 6 x - 12 = 0$ $x^2+2x-6x-12=0$

or $x (x + 2) - 6 (x + 2) = 0$ $x(x+2)-6(x+2)=0$

or $(x - 6) (x + 2) = 0$ $(x-6)(x+2)=0$

and hence $x = 6$ $x=6$ or $- 2$ $-2$

putting these values in $y = x + 2$ $y=x+2$ , we get $y = 8$ $y=8$ or $0$ $0$

Hence, we have two points $(6, 8)$ $(6,8)$ and $(- 2, 0)$ $(-2,0)$

graph{(y-x-2)(2x^2-y^2-8)=0 [-8, 12, -0.6, 9.4]}

What is the total number of points of intersection of the graphs of the equations 2x2−y2=8,y=x+22x^2-y^2=8, y=x+2?