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 $y=0$.
On
First, we find the derivative of the function, by multiplying by the equation by the original exponent $-2$, and subtracting $1$ off the exponent.
$$\frac{dy}{dx} = -4 \cdot x^{-3} = \frac{-4}{x^3}$$
At the point on the curve where $x=1, y=2$, and $\frac{dy}{dx} = -4$. The slope of the tangent at this point is thus $-4$, hence the slope of the normal is $\frac{1}{4}$.
Hence the intercepts are at $(0,1.75)=Q$ and $(-7,0) = P$. This means that $PQ$ has a length of $\sqrt{7^2 + 1.75^2} \approx 7.22$.
On
We have $\;y(1)=2$, $\;y'(1)=-\frac 4{x^3}\Bigl|_{x=1}=-4 $, so the equation of the tangent line at $(1,2)$ is $$ y-2 = -4(x-1)\iff 4x+y = 6? $$ Now the standard equation of a straightline, given its $x$-intercept $(a,0)$ and its $y$-intercept $(0,b)$ is $$\frac xa +\frac yb=1,$$ so rewriring the equation of the tangent line as $$\frac{2x}3+\frac y 6=1$$ instantly yields $$a=\frac 32,\quad b=6.$$ and thence the length $PQ$.
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.