My teacher told me to use the distance formula for this problem but I am not sure how to do it...
find the point (s) on the graph of y^2=6x closest to the point (5,0)
Let the point on the parabola which is closest to (5,0) be P(x,y)
let D be the distance between the two points
D^2 = (x-5)^2 + (y-0)^2
= (x-5)^2 + y^2
= (x-5)^2 + 6x
2D dD/dx = 2(x-5) + 6
for a minimum of D , dD/dx = 0
2(x-5) + 6 = 0
x-5 = -3
x = 2
then y^2 = 6(2)=12
y = ±√12 = ± 2√3
The two closest points are (2, 2√3) and (2, -2√3)
thank u for the help >_< this was so frustrating for me!
To solve this problem, you can use the distance formula to find the distance between the given point (5,0) and any arbitrary point on the graph y² = 6x. The point on the graph that is closest to (5,0) will be the one with the minimum distance.
Let's denote the coordinates of the arbitrary point on the graph as (x, y). The distance formula is:
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
Substituting the coordinates of the given point (5, 0) and the arbitrary point (x, y), we get:
Distance = √[(x - 5)² + (y - 0)²]
Since the arbitrary point (x, y) lies on the graph y² = 6x, we can substitute y² = 6x into the equation:
Distance = √[(x - 5)² + (y² - 6x)²]
Now, we need to find the values of x and y that minimize this expression.
To do that, we can use calculus. Take the derivative of the expression with respect to x, set it equal to zero, and solve for x. This will give us the x-coordinate of the closest point.
Differentiating the expression with respect to x, we have:
d(Distance)/dx = 2(x - 5) + 2(y² - 6x)(-6)
Setting this derivative equal to zero:
2(x - 5) + 2(y² - 6x)(-6) = 0
Simplifying:
2x - 10 - 12(y² - 6x) = 0
2x - 10 - 12y² + 72x = 0
74x - 12y² - 10 = 0
74x = 12y² + 10
x = (12y² + 10)/74
x = (6y² + 5)/37
Substituting this x value back into the equation y² = 6x, we get:
y² = 6((6y² + 5)/37)
37y² = 36y² + 30
y² = 30/37
Taking the square root of both sides:
y = ±√(30/37)
So, we have two possible y-values: y = √(30/37) and y = -√(30/37).
Substituting these y-values in the equation x = (6y² + 5)/37, we can find the corresponding x-values.
Therefore, the two points on the graph y² = 6x that are closest to (5,0) are (x, y) = ((6√(30/37)² + 5)/37, √(30/37)) and (x, y) = ((6(-√(30/37))² + 5)/37, -√(30/37)).
Note: Remember to simplify the expressions and consider both positive and negative square roots.