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 $O A$ is $\left(\frac{p}{q}\right)$ times as large as the area of the triangle $O A B$ (where $O$ is origin), then find the least value of $(p+q)$ where $p, q \in N$.

Answer 1

Given, $y=x^2$
Now, $\left.\frac{ dy }{ dx }\right|_{ x = a }=\left.2 x \right|_{ x = a }=2 a$
$\therefore$ Equation of tangent is $\left(y-a^2\right)=2 a(x-a)$
Put $y =0$, we get
$x = a -\frac{ a }{2}=\frac{ a }{2} \text {. }$
Also, equation of $OA$ is $y = ax$
$\text { Area } \left.=\int\limits_0^{ a }\left( ax - x ^2\right) dx = k \cdot \frac{ a }{2} \cdot \frac{ a ^2}{2} \Rightarrow \frac{ ax ^2}{2}-\frac{ x ^3}{3}\right]_0^{ a }=\frac{ ka ^3}{4} \Rightarrow \frac{ a ^3}{2}-\frac{ a ^3}{3}=\frac{ ka ^3}{4} \Rightarrow \frac{ a ^3}{6}=\frac{ ka ^3}{4}$
$\therefore k =\frac{2}{3} \equiv \frac{ p }{ q }($ Given $) \Rightarrow( p + q )_{\text {least }}=5$