A rather normal question

$Let\quad point\quad P(x,y)\quad so\quad equation\quad of\quad normal\quad is:\\ Y\quad -\quad y\quad =\quad -\frac { dx }{ dy } (X\quad -\quad x)\\ Now\quad for\quad Point\quad \quad Q(X,0)\quad \\ Y=0\quad \Longrightarrow \quad X=y\frac { dy }{ dx } \quad +\quad x\\ \Longrightarrow X\quad =4{ x }^{ 7 }+x\\ so\quad Q(4{ x }^{ 7 }+x,0)\\ So\quad Hight\quad of\quad \triangle POQ\quad =\quad y\quad =\quad { x }^{ 4 }\\ And\quad Base\quad OQ=4{ x }^{ 7 }+x\\ \\ Let\quad \\ E=\frac { Base }{ Hight } =\quad \frac { 4{ x }^{ 7 }+x }{ { x }^{ 4 } } =\quad 4{ x }^{ 3 }\quad +\quad \frac { 1 }{ { x }^{ 3 } } \\ \\ Now\quad For\quad To\quad Minimizes\quad E\quad we\quad use\quad Famous\\ inequality\quad for\quad Positive\quad variables\quad i.e\\ (To\quad avoid\quad Calculus\quad )\\ \\ AM\quad \ge \quad GM\\ \\ Take\quad Two\quad cases\quad on\quad 'x'\\ Case\quad 1:\quad x\quad >\quad 0\\ \\ \Longrightarrow \frac { 4{ x }^{ 3 }\quad +\quad \frac { 1 }{ { x }^{ 3 } } }{ 2 } \quad \ge \quad \sqrt { 4{ x }^{ 3 }.\quad \frac { 1 }{ { x }^{ 3 } } } \\ \\ \Longrightarrow E\quad \quad \ge \quad 4\\ { \Longrightarrow E }_{ min }=4\\ \\ Case\quad 2:\quad x\quad <\quad 0\\ \\ \frac { (-4{ x }^{ 3 })\quad +\quad (-\frac { 1 }{ { x }^{ 3 } } ) }{ 2 } \quad \ge \quad \sqrt { (-4{ x }^{ 3 }).(-\frac { 1 }{ { x }^{ 3 } } ) } \\ ({ -E }_{ min })\quad =4\\ \\ But\quad Physically\quad E\quad is\quad the\quad Ratio\quad of\quad sides\\ So\quad we\quad take\quad modulus\quad of\quad 'E'\\ \\ \Longrightarrow \quad \left| { -E }_{ min } \right| =\left| { E }_{ min } \right| =4\\ \\ So\quad required\quad Answer\quad is\quad 4\\$ .

Note: You don't need to solve These Two cases. Because By Logically thinking we conclude that There are Two such Triangles Possible for given condition. One in 1st quadrant and other in 2nd Quadrant. Both gives Same Minima so we have To calculate only For one case i.e. x>0 directly By AM-GM inequality for positive variables.

Let $P$ have coordinates $(p,p^{4})$ . (We can assume without loss of generality that $p \gt 0$ , since whatever result we find with $P$ in the first quadrant will be the same with $P$ in the second quadrant.) The slope of the tangent line to the curve $y = x^{4}$ at $P$ will be $4p^{3}$ , and thus the equation of the normal line to the curve at $P$ will be

$y - p^{4} = (-\frac{1}{4p^{3}})(x - p)$ .

To find point $Q$ we substitute $y = 0$ to find that that the $x$ -coordinate of $Q$ is $x = 4p^{7} + p$ . The ratio $R$ of $OQ$ to the height of $\Delta OPQ$ is then

$R = \dfrac{4p^{7} + p}{p^{4}} = 4p^{3} + \dfrac{1}{p^{3}}$ .

Now $\dfrac{dR}{dp} = 12p^{2} - \dfrac{3}{p^{4}} = 0$ when $p^{6} = \dfrac{1}{4} \Longrightarrow p^{3} = \dfrac{1}{2}$ .

This gives a minimum value for $R$ of $4*\dfrac{1}{2} + \dfrac{1}{\frac{1}{2}} = \boxed{4}$ .

(Note that we know this will be the minimum since $\frac{d^{2}R}{dp^{2}} \gt 0$ for $p \gt 0$ ).

That's interesting, $\Delta OPQ$ forms an isosceles triangle with base angles of $30$ degrees.