How do I find the normal of the curve $x^4 + 5y^4=21$? at $(2,1)$

Question

I know that the normal of a line is a line with a slope that is orthogonal to the line at a given point. The equation in question is $x^4 +5y^4 = 21$

In order to calculate the tangent line of the equation I simply differentiate and get the following.....

$4x^3dx + 20y^3dy = 0.$

And now I plug in my values from the given point $(2,1)$ and get....

$32\,dx + 20\,dy = 0$.

Thus my tangent line is $32(x-2) + 20(y-1) = 0$.

And this is were I am stuck. I have no clue on how to calculate the normal of the line. In my view, the normal line is $-\dfrac{1}{m}$ (with $m$ being the slope).

But when I plug in the answer $-\frac{1}{32} (x-2) - \frac1{20}(y-1) = 0$ it says its the wrong answer.

So how would one go about if they wanted to calculate the normal of this type of equation, because I have no clue.

Answer 1

You have worked out that the slope of the tangent is $-\frac{32}{20}$, and that the slope of the normal is the negative reciprocal of this or $\frac{20}{32}=\frac58$. The normal line, then, is the line with this slope passing through $(2,1)$: $y-y_0=m(x-x_0)$ gives $y-1=\frac58(x-2)$. The incorrect answer you derived is invalidated by a sign error; it is equivalent to $y-1=-\frac58(x-2)$.