Question

If $ \begin{bmatrix} 4a^2 & 4a & 1 \\ 4b^2 & 4b & 1 \\ 4c^2 & 4c & 1 \\ \end{bmatrix}
\begin{bmatrix} f(-1) \\ f(1) \\ f(2) \\ \end{bmatrix} =
\begin{bmatrix} 3a^2 + 3a \\ 3b^2 + 3b \\ 3c^2 + 3c \\ \end{bmatrix} $
f(x) is a quadratic function and its maximum value
occurs at a point V. A is a point of intersection of y = f (x)
with X-axis and point B is such th a t chord AB subtends
a right angle at V. Find the area enclosed by f(x) and
chord AB.

Answer 1

Given, $ \begin{bmatrix} 4a^2 & 4a & 1 \\ 4b^2 & 4b & 1 \\ 4c^2 & 4c & 1 \\ \end{bmatrix}
\begin{bmatrix} f(-1) \\ f(1) \\ f(2) \\ \end{bmatrix} =
\begin{bmatrix} 3a^2 + 3a \\ 3b^2 + 3b \\ 3c^2 + 3c \\ \end{bmatrix} $
$ \Rightarrow 4a^2 f(-1) + 4a f(1) + f(2) = 3a^2 + 3a ........(i) $
$ \ \ \ \ \ 4b^2\ f(-1) + 4\ b\ f(1) + f(2) = 3b^2 + 3b \ \ \ .........(ii) $
and $ \ \ \ \ \ \ \ 4c^2 f(-1) + 4c f(1) + f(2) = 3c^2 + 3c \ \ \ \ .........(iii) $
where, /(x) is quadratic expression given by,
$ \ \ \ \ \ \ f(x) = ax^2 + bx + c $ and Eqs. (i), (ii) and (iii).
$ \Rightarrow \ \ \ 4x^2 f(-1) + 4x f(1) + f(2) = 3x^2 + 3x $
or $ \{ 4 f(-1) - 3 \} x^2 + \{4f(1) - 3 \} x + f(2) = 0 \ \ \ \ .....(iv) $
As above equation has 3 roots a, b and c.
So, above equation is identity in x.
i.e. coefficients m ust be zero.
$ \Rightarrow \ \ \ \ \ f(-1) = \frac{3}{4} , f(1) = \frac{3}{4} , f (2) = 0 \ \ \ \ \ .....(v) $
$ \because \ \ \ \ \ \ f(x) = ax^2 + bx + c $
$ \therefore \ \ \ \ \ \ a = -\frac{1}{4} , b = 0 \ \ and \ \ c = 1, using Eq. (v) $
Thus, $ \ \ \ \ \ f(x) = \frac {(4 - x^2)}{4} $ shown as,
Let $ A (-2,0), B = (2t , -t^2 + 1) $
Since, AB subtends right angle at vertex V (0,1).
$ \Rightarrow \ \ \ \ 2\ \ 2t = -1 \ \ \Rightarrow t = 4 $
$ \therefore \ \ \ \ \ \ \ \ B(8, -15) $
So, equation of chord AB is $ y = \frac{-(3x + 6)}{2}. $
$ \therefore Required\ \ area = \bigg |\int \limits_{-2}^8 \bigg (\frac{(4 - x^2)}{4} + \frac{(3x + 6)}{2} \bigg) dx \bigg| $
$ \ \ \ \ \ \ \ \ \ = \bigg | \bigg (x - \frac{x^3}{12} + \frac{3x^2}{4} + 3x \bigg)_{-2}^{8} \bigg| $
$ \ \ \ \ \ \ \ \ \ = \bigg | \bigg [ 8 - \frac{128}{3} + 48 + 24 - \bigg ( -2 + \frac{2}{3} + 3 - 6 \bigg) \bigg] \bigg| $
$ \ \ \ \ \ \ = \frac{125}{3} sq\ \ units $