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Find the points P on the curve with equation $y=x^2-2$ such that the normal to the curve at P passes through the point (0,0).

I said

$y=x^2-2$

$\frac{dy}{dx} = 2x \rightarrow$ gradient of normal = $\frac{-1}{2x}$

Equation of normal: $y-0=\frac{-1}{2x}(x-0)\rightarrow y = \frac{-1}{2}$

At P, $x^2-2=\frac{-1}{2}\rightarrow$ coordinates are $\pm\sqrt\frac{3}{2},-\frac{1}{2}$

But the function $y=x^2-2$ has a minimum at $0,-2$ and I know the gradient here is 0 and the the normal to this passes through (0,0)but my calculation method does not pick this up.

Also, the gradient of the curve at (0,0) is 0 and that of the normal is $\infty$ so these cannot be multiplied together to get -1. Is the "rule" inappropriate here?

Steblo
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2 Answers2

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Your method doesn't pick up $(0,-2)$ as a valid answer because the gradient of the normal at $(0,-2)$ is not defined. The moment you write the gradient of the normal as $-\frac{1}{2x}$, it is automatically assumed that $x \ne 0$. So basically your solution works on the condition that $x \ne 0$.

In order to avoid missing solutions, you just have to separately check what happens in the case which you left out ($x=0$). And it is easy to check that the normal at $(0,-2)$ also satisfies the condition, which means that $(0,-2)$ is also a part of the solution set.

Ankit Saha
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$y=x^2-2$

$\frac{dy}{dx} = 2x$ so slope of tangent line at a point $(x_0, y_0)$ on the parabola is $2x_0$ and slope of the normal line is then $ - \frac{1}{2x_0}$.

But as the normal line also passes through the origin, you can simply equate the slopes,

$\frac{y_0 - 0}{x_0-0} = - \frac{1}{2x_0}$

As $(x_0, y_0)$ is on the parabola, $y_0 = x_0^2 - 2$ and so the above equation gives you values of $x_0$ where normal lines pass through origin.

I will leave details for you to work through and ask if you have any questions.

Math Lover
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