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 $OABC -2 \int\limits_{-1 /2}^{0}$ parabola

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