# How do you find all points on the curve #x^2 + xy + y^2 = 7# where the tangent line is parallel to the x-axis, and the point where the tangent line is parallel to the y-axis?

The tangent line is parallel to the

Solve (by substitution)

We could go through the solution, but the symmetry of the equation that we will get:

and the points on the curve at which the tangent is vertical are:

graph{x^2 + xy +y^2 =7 [-11.3, 11.2, -5.665, 5.585]}

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Using only middle school math I get

Tangents parallel to the x axis at:

Tangents parallel to the y axis at:

I glanced at Jim's answer, which looks like a nice, standard calculus treatment. But I couldn't help but feel sad for all the middle schoolers out there in Socratic land who want to find tangents of algebraic curves but are still years away from calculus.

Fortunately they can do these problems using only Algebra I.

This might be a bit complicated for a first example, but let's go with it. We write our curve as

Let's take

We expand, but we don't expand the difference terms

We said

We sorted the terms by degree, and we can experiment with approximations to

Let's just generate some approximations to

That's not particularly exciting, but it correctly tells us points near

Let's get more interesting and keep the linear terms.

When we set this to zero, we get the best linear approximation to

We can consider other approximations as well:

These are higher order tangents, ones that college math students hardly ever get to. We've already gone beyond college calculus.

There are more approximations, but I'm being warned this is getting long. Now that we learned how to do calculus using only Algebra I, let's do the problem.

We want to find the points where the tangent line is parallel to the

We found our tangent line at

Parallel to the

Since

Similarly parallel to the y axis means

Check.

How to check? Let's do an Alpha plot .

plot x^2+xy+y^2=7, x=-sqrt{7/3}, y=2 sqrt{7/3}, x=2sqrt{7/3}, y =-sqrt{7/3}

Looks good. Calculus on algebraic curves. Pretty good for middle school.

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To find points on the curve where the tangent line is parallel to the x-axis, we need to first find the derivative dy/dx of the curve equation with respect to x. Then, we set dy/dx equal to 0 and solve for the values of x. Substituting these x-values into the original equation will give us the corresponding y-values.

For points where the tangent line is parallel to the y-axis, we find the derivative dx/dy with respect to y. We set dx/dy equal to 0 and solve for the values of y. Substituting these y-values into the original equation will give us the corresponding x-values.

Finally, we obtain the coordinates of these points by combining the x and y values obtained from each step.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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