Question

Let 'c' be the constant number such that $c>1$. If the least area of the figure given by the line passing through the point $(1,c)$ with gradient ' $m$ ' and the parabola $y=x^2$ is 36 sq. units find the value of $\left(c^2+m^2\right)$.

Answer 1

Answer: 104

Equation of the line through $(1,c)$ is
$y-c=m(x-1)$
$y=mx+(c-m)$

$A=\underset{x_1}{\overset{x_2}{\int }}\left(mx+(c-m)-x^2\right)dx$

$=\left(x_2-x_1\right)\left[\frac{m\left(x_1+x_2\right)}{2}+(c-m)-\frac{x_1^2+x_2^2+x_1{x}_2}{3}\right]$
$=\sqrt{m^2+4(c-m)}\left[\frac{m^2}{2}+(c-m)-\frac{m^2-(m-c)}{3}\right]=\sqrt{m^2+4(c-m)}\left[\frac{m^2+4(c-m)}{6}\right]$
$=\frac{{\left[m^2+4(c-m)\right]}^{3/2}}{6}=\frac{{\left[m^2-4m+4c\right]}^{3/2}}{6}=\frac{{\left[(m-2{)}^2+4(c-1)\right]}^{3/2}}{6}$
$\therefore$ A is least if $m=2$
$A_{\text{least }}=\frac{[4(c-1){]}^{3/2}}{6}=\frac{2^3(c-1{)}^{3/2}}{6}=\frac{4}{3}\cdot (c-1{)}^{3/2}$
$\frac{4}{3}\cdot (c-1{)}^{3/2}=36\Rightarrow (c-1{)}^{3/2}=27\Rightarrow (c-1)={\left(3^3\right)}^{2/3}=9$
$c=10$
hence $c=10,m=2$
$\therefore c^2+m^2=100+4=104$
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