How do I find the point closest to (0,-7) for the hyperbola 7x^2-2y^2=20?

Question

So I know that this is a classic optimization problem where you’re looking for the minimum value and use the distance formula. I think the arithmetic is what is hanging me up. The answer I got for x is x= +/-sqrt(14/5) but I don’t think that’s correct. Plus, I need the y coordinate for each point.

can someone break it down for me, please?

7x^2-2y^2=20, closest to the point (0,-7)

Answer 1

Oh yay, I've thought of a better way.

I want the slope of the line from (0.-7) to the hyperbola to be perpendicular

to the slope of hyperbola which will be the case if the distance is minimal.

that is I want (y_h-(-7))/(x_h-0) which is line slope, equal to -(1/(dy/dx) which is negative

reciprocal inverse of dy/dx. Note implicit differentiation result is dy/dx=14x_h/4y_h for hyperbola

Not bad,, in equation (y+7)/x=(-4y)/14x and y= -98/18

plug in for x and x=+/-3.37

Answer 2

Sarah

Suppose the required point is (x,y)

Then if s = d² where d is the distance between (x,y) and (0,-7) then s = x² + (y+7)²

ds/dx = 2x + 2 (y+7) (dy/dx) *

And from the hyperbola equation

14x - 4y dy/dx = 0, so dy/dx = 14x/4y

When ds/dx = 0, 2x + 2 (y+7) (14x/4y) = 0

8xy + 28x (y+7) = 0

x = 0 or 8y + 28 (y+7) = 0 giving y = -5.44

x=0 is not a solution because there is no point on the hyperbola with x = 0

So y = -5.44, and substituting in to the hyperbolic equation gives x = +- sqrt (11.3) approx = +- 3.36 approx

So the points closest to (0,-7) are (-3.36, -5.44) and (3.36, 5.44) approx.

If you plot the hyperbola and the point on Desmos that looks right.

Answer 3

have to find min of f(x,y) = x² + (y + 7)² under conditions 7x² - 2y² - 20 = 0.

We create Lagrange function F(x,y) = x² + (y + 7)² - λ(7x² - 2y² - 20)

F_x = 2x - 14λx = 0, x = 0 or λ = 1/7

F_y = 2y + 14 + 4yλ = 0, 2y + 4/7y = - 14, y = - 14·7/18, y = -49/9;

F_λ= - 7x² + 2y² + 20 = 0; - 7x² + 2(-49/9)² = - 20; x² = 4021/567; x = ±√4021/567

f(±√4021/567, - 49/9) = 4021/567 + (- 49/9 + 7)² = 4021/567 + 196/81 = 5393/567

To show that this is minimum square of distance, let take some different point on hyperbola, point (2,2);

f(2,2) = 4 + 81 = 85 > 5393/567