Three points O(0,0),P(a,a^2),Q(-b,b^2),a0,b0, are on the parabola y=x^2.

Question

Three points $O(0,0),P(a,{a}^{2}),Q(-b,{b}^{2}),a>0,b>0$, are on the parabola $y={x}^{2}$. Let ${S}_{1}$ be the area of the region bounded by the line $PQ$ and the parabola, and ${S}_{2}$ be the area of the triangle $OPQ$. If the minimum value of $\frac{{S}_{1}}{{S}_{2}}$ is $\frac{m}{n},gcd(m,n)=1$, then $m+n$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 7

Accepted Answer

Equation of $PQ$

$\begin{gathered} y+b=\frac{{a}^{2}-{b}^{2}}{a+b}(x-{b}^{2}) \ \Rightarrow y=(a-b)x+ab \end{gathered}$

$\begin{gathered} {S}_{1}=\int _{-b}^{a}((a-b)x+ab-{x}^{2})dx \ ={(\frac{(a-b)}{2}{x}^{2}+abx-\frac{{x}^{3}}{3})}_{a}^{-b} \ =\frac{1}{6}(a+b{)}^{3} \end{gathered}$

${S}_{2}=\frac{1}{2}\begin{vmatrix} -b \quad  {b}^{2} \quad  1 \ 0 \quad  0 \quad  1 \ a \quad  {a}^{2} \quad  1 \end{vmatrix}=\frac{1}{2}ab(a+b)$

$\begin{gathered} \frac{{S}_{1}}{{S}_{2}} =\frac{\frac{1}{6}(a+b{)}^{3}}{\frac{1}{2}\cdot ab(a+b)} \ \ =\frac{1}{3}\frac{(a+b{)}^{2}}{ab}=\frac{1}{3}(\frac{a}{b}+\frac{b}{a}+2) \end{gathered}$

$\begin{gathered} \because \frac{b}{a}+\frac{a}{b}\ge 2 \ \Rightarrow {(\frac{{S}_{1}}{{S}_{2}})}_{min}=\frac{4}{3}=\frac{m}{n} \ \Rightarrow m+n=7 \end{gathered}$

Solutions · 1