Question

Consider the graph of $y=x^2$. Let $A$ be a point on the graph in the first quadrant. Let $B$ be the intersection point of the tangent on $y=x^2$ at the point $A$ and the $x$-axis. If the area of the figure surrounded by the graph of $y=x^2$ and the segment $OA$ is $\left(\frac{p}{q}\right)$ times as large as the area of the triangle $OAB$ (where $O$ is origin), then find the least value of $(p+q)$ where $p,q\in N$.

Answer 1

Answer: 5

Given, $y=x^2$
Now, ${\frac{dy}{dx}|}_{x=a}={2x|}_{x=a}=2a$
$\therefore$ Equation of tangent is $\left(y-a^2\right)=2a(x-a)$
Put $y=0$, we get
$x=a-\frac{a}{2}=\frac{a}{2}\text{. }$
Also, equation of $OA$ is $y=ax$
$\text{ Area }{=\underset{0}{\overset{a}{\int }}\left(ax-x^2\right)dx=k\cdot \frac{a}{2}\cdot \frac{a^2}{2}\Rightarrow \frac{a{x}^2}{2}-\frac{x^3}{3}]}_0^a=\frac{k{a}^3}{4}\Rightarrow \frac{a^3}{2}-\frac{a^3}{3}=\frac{k{a}^3}{4}\Rightarrow \frac{a^3}{6}=\frac{k{a}^3}{4}$
$\therefore k=\frac{2}{3}\equiv \frac{p}{q}($ Given $)\Rightarrow (p+q{)}_{\text{least }}=5$