Question

Let $A$ be the point on the curve $y=x^2$ in the first quadrant. Let $B$ be the point of intersection of the tangent to the curve $y=x^2$ at the point $A$ and the $x-axis$. If the area of the region bounded by the curve $y=x^2$ and the line segment $OA$ is $\frac{p}{q}$ times the area of the triangle $OAB,$ where $O$ is the origin, then the least positive integral value of $p+q$ is (Where $p$ and $q$ are co-promes numbers)

Answer 1

Parabola, $y=x^2............\left(1\right)(a=\frac{1}{4})$

Tangent to $\left(1\right)$ at $A(2at,a{t}^2)$ is $=>tx=y+a{t}^2..........\left(2\right)$

Point of intersection of $\left(2\right)$ with $X-axis$ is $B(at,0)$

Area of $△OAB=\frac{1}{2}\left|\begin{array}{cc}0& 0\\ at& 0\\ 2at& a{t}^2\\ 0& 0\end{array}\right|=\frac{1}{2}{a}^2{t}^3=\frac{t^3}{32}................\left(3\right)$

Equation of line $OA,$ $y=\frac{a{t}^2}{2at}x+0$

$=>y=\frac{t}{2}x.................\left(4\right)$

Area bounded between $\left(4\right)$ and $\left(1\right)=>A_{\left(say\right)}={\int }_0^{2at}\frac{t}{2}xdx-{\int }_0^{2at}{x}^{2}dx$

$=>A=t[\frac{x^2}{4}{]}_0^{2at}-[\frac{x^3}{3}{]}_0^{2at}$

$A=\frac{t^3}{16}-\frac{t^3}{(16{)}^3}(a=\frac{1}{4})$

$=>A=\frac{t^3}{13\left(8\right)}$

As given, $A=\frac{p}{q}$ (Area of $△OAB$)

$=>\frac{t^3}{3\left(8\right)}=\frac{p}{q}\left(\frac{t^3}{32}\right)$

$=>\frac{p}{q}=\frac{4}{3}$

$=>p=4$, $q=3$

Least value $(p+q)=4+3=7.$