Need help with calculus optimization problem

Question

Determine the point on the hyperbola −9x^2+2y^2=20 closest to the point (8, 0).

Give the x coordinate of each point: 

Give the positive y coordinate: 

Give the negative y coordinate:

Answer 1

graph the hyperbola and (8,0)

closest points on the hyperbola look about (4.42, 1.45) and (-4.42, 1.45)

whatever you get should be somewhere close to those two points

with x=1.45

and y=-4.42 and +4.42

-9x^2 +2y^2 = 20

is a hyperbola with two branches, symmetric about the x and y axes

with vertices on the y axis about (+/-3, 0)

2y^2 = 20+9x^2

y^2 = 10+4.5x^2

2yy' = 9x

y' = 9x/2y

slope of the perpendicular= -2y/9x,

one perpendicular line has positive slope, one has negative slope

find where the perpendicular lines through (8,0) intersects the hyperbola

x=16/11 = 1.454545....

y =+/- 4.41822 rounded off to nearest 5 decimals

distance squared = d^2

= (x-8)^2 + 10 +4.5x^2

= 5.5x^2 -16x +74

take the derivative, set = 0, solve for x

11x =16

x =16/11 = 1.454545...

y = +/- sqr(10- 4.5(16/11)^2)

= about +/- 4.41822

but someone else got y= sqr(2363)/11 = 4.41915,

so possible slight error above somewhere but very close

Answer 2

Minimize the distance, or equivalently the squared distance, between a point (x,y) on the hyperbola and the point (8,0):

d² = (x - 8)² + (y - 0)² = (x - 8)² + y²

Use the equation of the hyperbola to express y² in terms of x:

y² = 10 + (9/2)x²

Plug that in d² to obtain a function of a single variable x:

d² = (x - 8)² + 10 + (9/2)x²

Minimize d² with respect to x. That will give you a single value of x. Find the corresponding values of y from the equation for y².