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

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

$A=\int\limits_{x_1}^{x_2}\left(m x+(c-m)-x^2\right) d x$

$=\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] $
$=\sqrt{m^2+4(c-m)}\left[\frac{m^2}{2}+(c-m)-\frac{m^2-(m-c)}{3}\right]=\sqrt{m^2+4(c-m)}\left[\frac{m^2+4(c-m)}{6}\right]$
$=\frac{\left[m^2+4(c-m)\right]^{3 / 2}}{6}=\frac{\left[m^2-4 m+4 c\right]^{3 / 2}}{6}=\frac{\left[(m-2)^2+4(c-1)\right]^{3 / 2}}{6}$
$\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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