Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Calculating the equation of the curve.
y-b=a(x-4)² ----- (1)
At coordinate (8, 0)
-b=16a
Therefore;
a = -b/16
Sub. a in (1) to get the equation of the curve.
y-b=
(-b/16)(x-4)²
It implies;
y = (-bx²/16)+(bx/2) ----- (2)
Calculating b.
The area of the curve at point (8, 0) = 1280/27
It implies;
(-32b/3)+16b=1280/27
b = 1280/(16*9)
b = 80/9
Therefore sub. b in (2).
y = (-(80/9)x²/16)+((80/9)x/2)
y = (-5x²/9) + (40x/9)
Calculating y, when XQ = 6.
y = (-5*36)/9 + (40*6)/9
y = 20/3
Calculating the point of intersection of forward slant length Q and backward slant length P.
For P, the equation is;
y = (-20x+160)/9
For Q, the equation is;
y = (10x)/9
Therefore, solving both equations simultaneously;
x = 16/3
y = 160/27
Calculating Area S.
It is;
Area under the curve at point (6, 0) and (8, 0) + Area trapezoid with parallel sides (20/3) units and (160/27) units, and height ⅔ units - Area triangle with height (160/27) units and base (8/3) units
(200/27) + ½((20/3)+(160/27))*(2/3) - ½*(160/27)*(8/3)
= (200/27) + (340/81) - (640/81)
= (200/27) + (-300/81)
= (200/27) - (100/27)
= 100/27 space units.
Calculating Area R.
It is;
Area under the curve at point (4, 0) and (8, 0) - Area triangle with height (80/9) units and base 4 units - Area S.
(640/27) - ½*(80/9)*(4) - (100/27)
= (640/27) - (160/9) - (100/27)
= 60/27
= 20/9 space units.
Therefore;
Area R ÷ Area S is;
(20/9) ÷ (100/27)
= 60/100
= 3/5
