Plotting a curve

[burgerandcheese]

Question: The equation x² - 6xy + 25y² = 16 represents an ellipse with its centre at the origin. What ranges of values of x and y would you need in order to plot the whole of the curve on a computer screen?

Well one suggested method is by rewriting the equation as a quadratic in x and then set the discriminant >= 0 and doing the same for y

so -1 <= y <= 1
-5 <= x <= 5

I don't understand why this works? What I know is that the discriminant only tells us the types of roots of a quadratic equation.
Thank you in advance

 

[MarkFL]

When we consider the curve as a quadratic in \(x\) we get upon requiring the discriminant to be non-negative (so that only real roots are considered):

[MATH]36y^2-4(25y^2-16)\ge0[/MATH]
[MATH]1-y^2\ge0[/MATH]
And this implies:

[MATH]1\le y\le1[/MATH]
What we have done is find the range of real \(y\)-values. The same thing will work when we consider the curve a quadratic in \(y\), we will get the range of real \(x\)-values.

 

[Dr.Peterson]

[MATH][/MATH]I don't understand why this works? What I know is that the discriminant only tells us the types of roots of a quadratic equation.

I see you've just been answered with regard to y; let's look at the domain, thinking specifically about why the method works; that is, what we are really doing.

As a quadratic equation in y, we have [MATH]25y^2 + (-6x)y + (x^2 - 16) = 0[/MATH]. We are looking for the values of x for which there will be a real value of y that satisfies the equation. So we find the discriminant, [MATH](-6x)^2 - 4(25)(x^2 - 16) = 36x^2 - 100x^2 + 1600 = 64(1 - 25x^2)[/MATH]. This is non-negative for [MATH]-5 \le x \le 5[/MATH], so for these values of x there will be a point (x,y) on the graph. And that is what we needed.