Question

A variable line $y=l\left(x\right)$ intersects the parabola $y=x^2$ at points $P$ and $Q$ whose $x$-coordinates are $\alpha$ and $\beta$ respectively with $\alpha <\beta$. The area of the figure enclosed by the segment $PQ$ and the parabola is always equal to $\frac{4}{3}$. The variable segment $PQ$ has its middle point as $M$. Then the value of $(\beta -\alpha )$ is greater than (or) equal to-

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

Answer: (A), (B)

The correct options are
A $1$
B $2$

$y=x^2$
Let $P(\alpha ,{\alpha }^2)$ and $Q(\beta ,{\beta }^2)$ be the points of intersection of line and parabola
$PQ\to (y-{\alpha }^2)=(\beta +\alpha )(x-\alpha )y=(\beta +\alpha )x-\alpha \beta$
Area $=\underset{\alpha }{\overset{\beta }{\int }}\left(\right((\alpha +\beta )x-\alpha \beta )-x^2)dx$
$\frac{4}{3}={\left(\right(\alpha +\beta )\frac{x^2}{2}-\alpha \beta x-\frac{x^3}{3})}_{\alpha }^{\beta }$
$\Rightarrow (\beta -\alpha {)}^3=8\beta -\alpha =2$