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find the points on the parabola $2x+y^2=0$ closest to the point $(-3,0)$

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Find Coordinate on Unit Circle

Full solution

Q. find the points on the parabola

2x+y^2=0

closest to the point

(-3,0)

Use Distance Formula: We need to minimize the distance from the point $(-3,0)$ to any point $(x,y)$ on the parabola $2x + y^2 = 0$ . $\newline$ Use the distance formula: $D = \sqrt{(x - (-3))^2 + (y - 0)^2}$ .
Simplify Formula: Simplify the distance formula: $D = \sqrt{(x + 3)^2 + y^2}$ .
Minimize Distance: To find the minimum distance, we can minimize $D^2$ to avoid the square root: $D^2 = (x + 3)^2 + y^2$ .
Substitute $y^2$ : Substitute $y^2$ from the parabola equation into $D^2$ : $D^2 = (x + 3)^2 + (-2x)$ .
Expand and Simplify: Expand and simplify: $D^2 = x^2 + 6x + 9 - 2x$ .
Combine Like Terms: Combine like terms: $D^2 = x^2 + 4x + 9$ .
Take Derivative: To find the minimum, take the derivative and set it to zero: $\frac{d(D^2)}{dx} = 2x + 4$ .
Set Derivative to Zero: Set the derivative equal to zero and solve for $x$ : $2x + 4 = 0$ .
Solve for x: Solve for x: $x = -\frac{4}{2}$ .
Substitute Back: Solve for $x$ : $x = -2$ .
Find $y$ : Substitute $x$ back into the parabola equation to find $y$ : $2(-2) + y^2 = 0$ .
Closest Points: Solve for $y^2$ : $y^2 = 4$ .
Closest Points: Solve for $y^2$ : $y^2 = 4$ . Find $y$ : $y = \pm 2$ .
Closest Points: Solve for $y^2$ : $y^2 = 4$ . Find $y$ : $y = \pm 2$ . The closest points on the parabola are $(-2,2)$ and $(-2,-2)$ .

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