Four points A, B, C and D lie in that order on the parabola y = ax 2 + bx + c. The coordinates of A, B and D are known as: A(–2, 3), B(–1, 1), D(2, 7). Find the coordinates of C for which the area of the quadrilateral ABCD is the greatest.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Let the coordinates of C be (x, y). Since points A, B and D lie on y = ax 2 + bx + c. Therefore,
3 = 4a –2b + c
1 = a – b + c
7 = 4a + 2b + c
Solving these equations, we get
a = 1, b = 1, c = 1
∴ y = x 2 + x + 1 ... (i)
Clearly, area quadrilateral ABCD will be maximum, if the area of Δ BCD is maximum. Let P denote the area of Δ BCD. Then,
P = 
= 
(2x – y + 3)
= 
(2x –x 2 – x –1 + 3) [Using (i)]
= 
(–x 2 + x + 2)
∴ 
= 
(–2x + 1) and, 
= –3 < 0
For maximum or minimum values of P, we must have
= 0 ⇒ –2x + 1 = 0 ⇒ x = 
Clearly, 
< 0 for all x.
Hence, P is maximum when x = 
Putting x = 
in (i), we get y = 
Hence, the coordinates of C are (1/2, 7/4).
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems