A line is drawn normal to the curve $y=\frac{2}{x^2}$ at the point on the curve where x=1. This line cuts the x-axis at P and the y-axis at Q. What is the length of PQ?
When $x=1$, $y=2$. I don't know how to find the points P and Q, but I know that for Q $x=0$ and for P $y=0$.
Best Answer
The slope of the tangent is $y'= - \frac{4}{x^3} |_{x=1} = -4$
The equation of the tangent is $y-2 = -4(x-1) \implies y+4x = 6$
At the $x-$axis $y=0$ and at the $y-$ axis, $x=0$
From this find $P(a,0) \ , Q(0,b)$
The length will then be $\sqrt{a^2+b^2}$
EDIT: The slope the normal line is $m = - 1/y' = 1/4$
So, the equation of the normal line is $y-2 = \frac{1}{4}(x-1) \implies 4y - 8 = x - 1 \implies \color{red}{4y = x+7}$
The remaining part is the same.