If $\left[\begin{array}{ccc}4 a^2 & 4 a & 1 \ 4 b^2 & 4 b & 1 \ 4 c^2 & 4 c & 1\end{array} ight]\

Question

If $\left[\begin{array}{ccc}4 a^2 & 4 a & 1 \ 4 b^2 & 4 b & 1 \ 4 c^2 & 4 c & 1\end{array}ight]\left[\begin{array}{c}f(-1) \ f(1) \ f(2)\end{array}ight]=\left[\begin{array}{l}3 a^2+3 a \ 3 b^2+3 b \ 3 c^2+3 c\end{array}ight]$ where $f(x)$ is a quadratic function and $f(x)=a x^2+b x+c$ whose maximum value occurs at a point $V$ say $(\alpha, \beta)$. Let $A$ be the point of intersection of $y=f(x)$ with negative $x$-axis, say $(p, o)$ and point $B$ is such that the chord AB subtends a right angle at V . Let B be $(\mathrm{r}, \mathrm{s})$. Let $\Delta$ be the area enclosed by $y=f(x)$ and the chord AB . Then given $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are distinct real numbers.

Professor PC Thomas Classes · Answer

$ \begin{aligned} & 4 a^2 f(-1)+4 a f(1)+f(2)=3 a^2+3 a \ & \frac{4 b^2 f(-1)+4 b f(1)+f(2)=3 b^2+3 b}{4 f(-1)\left(a^2-b^2ight)+4 f(1)(a-b)} \ & =3\left(a^2-b^2ight)+3(a-b) \ & 4 f(-1)(a+b)+4 f(1)=3(a+b)+3 \ & 4 f(-1)=3 \quad 4 f(1)=3 \ & f(-1)=\frac{3}{4}, \quad f(1)=\frac{3}{4} \ & f(2)=0 \ & f(2)=0 \ & 4 a+2 b+c=0 \ & a-b+c=\frac{3}{4} \ & a+b+c=\frac{3}{4} \ & \frac{2 b=0 \Rightarrow b=0}{4 a+c=0} \ & a+c=\frac{3}{4} \ & 3 a=-\frac{3}{4} \Rightarrow a=-\frac{1}{4} \ & c=\frac{3}{4}+\frac{1}{4}=1 \end{aligned} $ $\begin{aligned} & f(x)=-\frac{1}{4} x^2+1 \ & (\alpha, \beta)=(0,1) \ & A=(-2,0) \Rightarrow P=-2 \ & -\frac{\frac{x^2}{4}}{x} 	imes \frac{1}{2}=-1 \Rightarrow x=8 \quad B(8,-15)\end{aligned}$ $$ \begin{aligned} & y-0=-\frac{3}{2}(x+2) \ & \Rightarrow 2 y=-3 x-6 \Rightarrow y=-\frac{3}{2} x-3 \ & \Rightarrow 3 x+2 y+6=0 \ & y_1-y_2=-\frac{x^2}{4}+1+\frac{3}{2} x+3=-\frac{x^2}{4}+\frac{3 x}{2}+4 \ & \Delta=\frac{\left(\frac{9}{4}-4 	imes-\frac{1}{4} 	imes 4ight)^{3 / 2}}{6 	imes \frac{1}{16}}=\frac{125}{8} 	imes \frac{8}{3} \end{aligned} $$