a car is traveling along a highway shaped like a parabola with its vertex at the origin. the car starts at 100 miles north and 100 miles west. there is a statue at 100 miles east and 50 miles north. at what point will the cars headlights hit the statue?

Answer 1

y = a x^2
100 = a(10,000)
a = .01
so our parabola is y = .01 x^2

now a tangent to that parabola must pass through (100,50)
slope = .02 xp at parabola
y = (.02xp)x + b
50 = 2 xp + b
b = 50-2xp
but
yp = .01 xp^2
and
yp =.02xp^2 +50-2xp
.01 xp^2 = .02xp^2 + 50 - 2 xp
.01 xp^2 -2xp +50 = 0
xp^2-200xp+5000 = 0
xp = 100 +/- 50sqrt2
so about 70
if xp = 70
yp = .01x^2 = 49

Answer 2

let the equation of the parabola be
y = ax^2 , with (-100,100) on it
100 = a(-100)^2
100 = 10,000a
a = 100/10000 = 1/100

so y = (1/100)x^2
The car's headlights will hit the point (100,50)
when the light-line is tangent to the parabola at the point (a,b)
dy/dx = (1/50)x
at (a,b)
dy/dx = a/50
also b = a^2/100

slope of tangent:
(b-50)/(a-100)

so (b-50)/(a-100) = a/50
b - 50 = (a/50)(a-100)
a^2/100 = a^2/50 - 2a + 50
times 100
a^2 = 2a^2 - 200a + 5000
0 = a^2 - 200a + 5000
a = (200 ± √20000)/2
= (200 ± 100√2)/2
= 100± 50√2
= appr 29.29 miles or 170.71 miles

if a = 29.29 , b = 8.57
if a = 170.71, b = 291.42

As the car approaches from the left, the SUPER-LIGHT will hit the statue when the car is
29.3 miles east and 8.6 miles north

This is silly question. The car's headlights would have to shine over 80 miles !!!

Answer 3

xp = 100 +/- 50sqrt2
so about 30 (NOT 70 AS I SAID)
if xp = 30
yp = .01x^2 = 9