write the standard form of the equation of the parabola with the following A) Center (0,0) and r=7 B) Center (-3,2) and r =5

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You should remember what is the standard form of equation of the circle such that:

`(x-h)^2+(y-k)^2=r^2`

Notice that (h,k) represents the center of circle and r represents the radius.

Since the problem provides the coordinates of the center and the value of radius, you just need to substitute these values in...

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You should remember what is the standard form of equation of the circle such that:

`(x-h)^2+(y-k)^2=r^2`

Notice that (h,k) represents the center of circle and r represents the radius.

Since the problem provides the coordinates of the center and the value of radius, you just need to substitute these values in standard form of equation of circle such that:

`(x-(-3))^2+(y-2)^2=5^2`

`(x+3)^2+(y-2)^2=25`

Hence, evaluating the standard form of equation of the circle, whose center is at (-3,2) and its radius is of 5, yields `(x+3)^2+(y-2)^2=25.`

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A parabola can be written as

`(y-y_0) = (x-x_0)^2/(2r)`

where `(x_0,y_0)` is it's centre, or vertex and `r` is the radius of curvature at the centre (twice the focal length).

In standard form a parabola is written as

`y = ax^2 + bx + c`

A) `(y-0) = (x-0)^2/7`  `implies`  `y = x^2/7`

B) `(y-2) = (x-(-3))^2/5`  `implies`  `y = (x+3)^2/5 + 2`

 `implies `  `y = 1/5(x^2 + 6x + 9) + 2`

`implies`  `y = 1/5x^2 + 6/5x + (9+10)/5 = 1/5x^2 + 6/5x + 19/5` ` `

A) y = x^2/7

B) y = 1/5x^2 + 6/5x+ 19/5

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