Question

The maximum possible area bounded by the parabola $y=x^2+x+10$ and a chord of the parabola of lenghth $1$ is

• A. $\frac{1}{12}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

We have,
$y=x^2+x+10$
$\Rightarrow x^2+x+\frac{1}{4}$
$=y-10+\frac{1}{4}$
$\Rightarrow {\left(x+\frac{1}{2}\right)}^2=\left(y-\frac{39}{4}\right)$
Latusrectum of parabola is 1
$\therefore$ Length of chord is also 1
Area is maximum when chord is latusrectum
Area of shaded region =
Area of rectangle $OABC-2\underset{-1/2}{\overset{0}{\int }}$ parabola

$=10-2\underset{\frac{-1}{2}}{\overset{0}{\int }}\left(x^2+x+10\right)dx$
$=10-2{\left[\frac{x^3}{3}+\frac{x^2}{2}+10\right]}_{\frac{-1}{2}}^0$
$=10-2\left[\frac{-1}{24}+\frac{1}{8}-\frac{10}{2}\right]$
$=\frac{1}{6}$