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I have this problem: "Find a number k such that the line $x + y = k$ is normal to the curve $y = x^{2}$

I did it like this:

$y' = 2x \Rightarrow y'(a) = 2a$

$y(a) = a^{2}$

So I put this into the formula for a tangent to an equation $y-y_0 = a(x-x_0)$

And got: $y - a^{2} = 2a(x-a)$, and then I found the tangent that is normal to that one:

$y - a^{2} = \frac{-1}{2a}(x-a) \Leftrightarrow y = \frac{-1}{2a}x + \frac{1}{2} + a^{2}$

Then I just added x to both sides and said

$ x + y = \frac{-1}{2a}x + \frac{1}{2} + a^{2} + x $

So $k \text{ is } \frac{-1}{2a}x + \frac{1}{2} + a^{2} + x$

... but my classmate said it's wrong and also im skeptical to my answer since the problem asks for a number k

Anyone know how to interepret this problem? Thanks in advance

A. Lindberg
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1 Answers1

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If a tangent at the point $(a,a^2)$ has a gradient $2a$ then its normal has a gradient $-\frac{1}{2a}$, as you have found.

You want the normal to be of the form $y=-x+k$ i.e. with gradient $-1$, so you want $a=\frac12$ and thus the normal passes through the point $(\frac12,\frac14)$, so $k=x+y=\frac34$.

This is what you would get from your final expression with $a=\frac12$.

Henry
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