Question

Find the points of intersection of parabola `y=-2x^2-3x-4` and the circle `x^2+y^2+2x-2y-7=0`?

Answer 1

The two do not intersect.

Explanation:

Let us consider the parabola `y=-2x^2-3x-4`, which in vertex form is `y=-2(x+3/4)^2-23/8`. As coefficient of `x^2` is negative we have a maxima at `(-3/4,-23/8)` or `(-0.75,-2.875)`.

The circle `x^2+y^2+2x-2y-7=0` can be written as `(x+1)^2+(y-1)^2=3^2` i.e. its center is `(-1,1)` and radius is `3`. and circle lies between `y=-2` and `y=4`. But parabola has a maxima at `y=-2.875`.

Hence, parabola lies below circle and they do not intersect each other.

Further, observe that the distance between `(-1,1)` and vertex of parabola `(-3/4,-23/8)` is

`sqrt((-1+3/4)^2+(1-(-23/8))^2)`

= `sqrt(1/16+961/64)=1/8sqrt965=31.064/8=3.883`

which is greater than `3`. Hence the two do not intersect.

graph{(y+2x^2+3x+4)(y+2)(y-4)(8y+23)(x^2+y^2+2x-2y-7)=0 [-10, 10, -5, 5]}