Respuesta :
circle: x2+ y2 = 2
parabola: y = 2x2 – 3
The standard form of equation of circle is:
( x – h )^2 + ( y – k )^2 = r^2
We are given the equation x^2 + y^2 = 2.
Where h and k are the x and y coordinates of the center. Basing on the standard equation, the center is at (0 , 0) and r = [tex] \sqrt{2} [/tex]
The standard form of equation of parabola with the axis of symmetry parallel to the y-axis is:
(x - h)^2 = 4p (y - k)
Rearranging the given equation into this form:
x^2 = (½) y + 3/2
x^2 = ½ (y + 3)
Since 4p is positive value, therefore the parabola is opening up. The vertex is located at (0 , -3).
Answer: circle with its center at (0, 0) and a radius of [tex] \sqrt{2} [/tex]; parabola opening up with its vertex at (0, –3)
The equation [tex]x^2+ y^2 = 2[/tex] represents a circle of radius [tex]\sqrt 2[/tex] and center (0,0), and the equation [tex]y=2x^2-3[/tex] represents a parabola with vertex (0,-3) and it opens upward.
Given information:
The given equations are:
[tex]x^2+ y^2 = 2\\y = 2x^2 - 3[/tex]
Let's talk about the equations one by one.
The first equation is [tex]x^2+ y^2 = 2[/tex].
The given equation represents a circle with its center at (0,0) and radius [tex]\sqrt2[/tex]. The equation can also be represented as,
[tex]x^2+ y^2 = 2\\(x-0)^2+(y-0)^2=(\sqrt2)^2[/tex]
Now, the second equation is [tex]y=2x^2-3[/tex]. It is the equation of a parabola.
The given equation can be written as,
[tex]y=2x^2-3\\y-(-3)=2(x-0)^2[/tex]
So, the vertex of the parabola is (0,-3) and it opens upwards. The opening is upwards because the coefficients of x and y are positive and y is related to x squared.
Therefore, the equation [tex]x^2+ y^2 = 2[/tex] represents a circle of radius [tex]\sqrt 2[/tex] and center (0,0), and the equation [tex]y=2x^2-3[/tex] represents a parabola with vertex (0,-3) and it opens upward.
See the image attached.
For more details, refer to the link:
https://brainly.com/question/15136456
