Question

The curve $f\left(x\right)=A{x}^2+Bx+C$ passes through the point (1, 3) and line $4x+y=8$ is tangent to it at the point (2, 0). The area enclosed by $y=f\left(x\right),$ the tangent line and the y-axis is

• A. $\frac{4}{3}$
• B. $\frac{8}{3}$
• C. $\frac{16}{3}$
• D. $\frac{32}{3}$

Answer 1

The correct option is B $\frac{8}{3}$

Given curve is $y=f\left(x\right)=A{x}^2+Bx+C.........\left(i\right)$

It passes through $(1,3)$
$\therefore 3=A+B+C..........\left(ii\right)$

Point $(2,0)$ also lie on the curve $\left(i\right)$
$\therefore 0=4A+2B+C.........\left(iii\right)$

$-C=4A+2B$
from eq (ii)

$3=A+B-4A-2B$

$B=3(1-A)........\left(v\right)$

Slope of tangent is $-4$

$\therefore -4=4A+B.........\left(iv\right)$

$\therefore$ from $\left(ii\right),\left(iii\right)$ & $\left(iv\right)$ and $\left(v\right)$ we get

$A=-1,B=0,C=4$

Thus, the required curve is $y=-x^2+4$
Hence required area $=$ area of $\Delta OAB-{\int }_0^2(-x^2+4)dx=8-(-\frac{8}{3}+8)=\frac{8}{3}$