Given a straight line passing through (0,−2) and (23,0). The distance from the parabola y=x2−1 to that line is....
Explanation
Create the equation of the line from the points it passes through, namely (0,−2) and (23,0)
y2−y1y−y1=x2−x1x−x1
0−(−2)y−(−2)=23−0x−0
2y+2=32x
3y+6=4x
−4x+3y+6=0
We cannot directly find the distance from a point to the line. Find the point whose position is closest to the line. Assume the point is (a,b) so that
y=x2−1
b=a2−1
The point becomes (a,a2−1) as shown in the following illustration
Parabola and Line Visualization
The graph shows the parabola y=x2−1 and the line −4x+3y+6=0 with the closest point (32,−95).
We need to determine the distance from point (a,a2−1) to the line −4x+3y+6=0
distance=(−4)2+32∣−4x+3y+6∣
f(a)=5∣−4a+3(a2−1)+6∣
f(a)=5∣3a2−4a+3∣
f(a)=51(3a2−4a+3)
The condition for minimum value is f′(x)=0
f′(a)=51(6a−4)
f′(a)=0
51(6a−4)=0
6a=4
a=64=32
Distance between line and parabola when a=32
f(a)=51(3a2−4a+3)
f(a)=51(3(32)2−4(32)+3)
f(a)=51(34−38+39)
f(a)=51(35)=31