A hyperbola analytic geometry problem

[Ognjen]

I struggle to understand the solution to an analytic geometry problem from my workbook, concerning a chord going through a hyperbola.

I don't know if I will translate it correctly ( since the original problem is in Serbian ), as I may not know the English terms for the concepts in question, but here's an attempt:

"Determine the "geometric placement" ( I assume this is an equivalent to a locus in English, but I don't recognize the Serbian term to begin with, so I may not have translated it properly ) of the "middles of chords" passing through a hyperbola:

[math]x^2 - 4y^2 = 16[/math]
..., so that they ( the chords ) make an angle of [math]\pi/4[/math]
The solution first requires to write the general equation of a line, taking into account the given slope ( so basically y = x + n ), and put it into the hyperbola equation.
I get:
[math]3x^2 + 8xn + 4n^2 + 16 = 0[/math]
Since the line has to contain a CHORD of the hyperbola, it should supposedly intersect it at 2 distinct points. This means that the given quadratic equation ( by x ) should have the discriminant OVER 0, so that it gives 2 different solutions, determining 2 different points. HOWEVER, instead of this rationale that I just expounded upon, the official solution just says: "...since the solutions have to be real numbers, the discriminant has to be EQUAL OR ABOVE 0 ). So what they get is that
[math]|n| >= 2\sqrt(3)[/math]and my solution for this differentiates in that for me its JUST above 0, not "or equal". Back on track ( since this seems to be the less confusing part ):

The solution now suggests to devise a point S(x,y), so that it is placed in the middle of the chords ( in accordance with the guidelines of the problem ). So:
[math]S(x,y) = S((x_1 + x_2)/2, (y_1 + y_2)/2))[/math]with coordinates x1 y1 and x2 y2 belonging to 2 distinct points of intersection of the lines containing the chords. That much is clear.

Then, by using Vieta's formulas on the initial quadratic, we have [math]x_1 + x_1 = -8n/3[/math]Replacing (y1 + y2)/2 for y, and (x1 + x2)/2 for x in the line equation from the very beginning, we now have:
[math](y_1 + y_2)/2 = (x_1 + x_2) / 2 + n[/math][math]y_1 + y_2 = -8n/3 + 2n[/math][math]y_1 + y_2 = -2n/3[/math]
Now, we can find the full line equation by "getting rid of the n":
[math]((y_1 + y_2)/2) / ((x_1 + x_2)/2) = 1/4[/math][math]y/x = 1/4[/math][math]y = 1/4*x[/math]
The solution now requires finding the coordinates of the 2 points, by plugging the newly found y into the hyperbola equation. So I get [math]x = +-8\sqrt(3)/3[/math] and [math]y = +- 2\sqrt(3)/3[/math]
Now, what exactly the solution is here ? Based on the text of the problem, I would now find the middle point between the 2 points I just found, or perhaps the distance between them ( I don't know what "geometric placement of the middles of chords" really is, so this much would be intuitive ) and call it a day.
HOWEVER, it now says the solution is:
y = 1/4 * x, FOR [math]|x| >= 8\sqrt(3)/3[/math]
Does this simply mean the solution is the part of the y=1/4*x line that passes through the two points ? If that's the case, what use did we have from the confusing n>= z requirements ? Why even bother determining it, if the n is just gonna disappear ? I don't understand at all.

Thank you for your effort in advance.

 

[Dr.Peterson]

Since the line has to contain a CHORD of the hyperbola, it should supposedly intersect it at 2 distinct points. This means that the given quadratic equation ( by x ) should have the discriminant OVER 0, so that it gives 2 different solutions, determining 2 different points. HOWEVER, instead of this rationale that I just expounded upon, the official solution just says: "...since the solutions have to be real numbers, the discriminant has to be EQUAL OR ABOVE 0 ). So what they get is that
[math]|n| >= 2\sqrt(3)[/math]and my solution for this differentiates in that for me its JUST above 0, not "or equal". Back on track ( since this seems to be the less confusing part ):

The only difference is that they are including the end point, where the "chord" is really a single point (of tangency), whereas you, perhaps more correctly, exclude it. Their locus will be a pair of rays, yours will lack the end points.

The solution now suggests to devise a point S(x,y), so that it is placed in the middle of the chords ( in accordance with the guidelines of the problem ). So:
[math]S(x,y) = S((x_1 + x_2)/2, (y_1 + y_2)/2))[/math]with coordinates x1 y1 and x2 y2 belonging to 2 distinct points of intersection of the lines containing the chords. That much is clear.

Then, by using Vieta's formulas on the initial quadratic, we have [math]x_1 + x_1 = -8n/3[/math]Replacing (y1 + y2)/2 for y, and (x1 + x2)/2 for x in the line equation from the very beginning, we now have:
[math](y_1 + y_2)/2 = (x_1 + x_2) / 2 + n[/math][math]y_1 + y_2 = -8n/3 + 2n[/math][math]y_1 + y_2 = -2n/3[/math]
Now, we can find the full line equation by "getting rid of the n":
[math]((y_1 + y_2)/2) / ((x_1 + x_2)/2) = 1/4[/math][math]y/x = 1/4[/math][math]y = 1/4*x[/math]

I would not have done all this work, but just used the quadratic formula to find the x-coordinates of the two points of intersection, averaged them, and then put that into the equation of the line to find the y-coordinate of the midpoint of the chord.

What they have done is to find the x-coordinate using the Vieta trick, then find the average of the y-coordinates using another trick, and then eliminate n. The solution requires nothing more: you have the equation of the locus, and the restriction due to requiring a chord.

The solution now requires finding the coordinates of the 2 points, by plugging the newly found y into the hyperbola equation. So I get [math]x = +-8\sqrt(3)/3[/math] and [math]y = +- 2\sqrt(3)/3[/math]
Now, what exactly the solution is here ? Based on the text of the problem, I would now find the middle point between the 2 points I just found, or perhaps the distance between them ( I don't know what "geometric placement of the middles of chords" really is, so this much would be intuitive ) and call it a day.
HOWEVER, it now says the solution is:
y = 1/4 * x, FOR [math]|x| >= 8\sqrt(3)/3[/math]

I don't see why you say there is still a need to find the y-coordinates of the end points of the chord, and thereby both coordinates of the end point of the locus (ray). The solution is, as you say, "y = 1/4 * x, FOR [imath]|x| >= 8\sqrt(3)/3[/imath]", which doesn't require that.

But, again, the more direct approach is, as you say, not to use all the trickery to find the averages, but just find the x-coordinates of the endpoints of the chord, average them, find y from the equation of the line, and then state the answer.

The "geometric placement of the middles of chords" is the locus, namely the equation satisfied by the midpoint of every such chord.

Does this simply mean the solution is the part of the y=1/4*x line that passes through the two points ? If that's the case, what use did we have from the confusing n>= z requirements ? Why even bother determining it, if the n is just gonna disappear ? I don't understand at all.

The restriction on n for forming a chord is needed to determine which part of the line y = x/4 forms the locus. That could instead have been done using the discriminant.

Here is a graph of the hyperbola (red), a chord (blue), its midpoint (green), and the locus (green):

The locus consists of all places the green point could be.

Does that make things any clearer?

 

[Ognjen]

The only difference is that they are including the end point, where the "chord" is really a single point (of tangency), whereas you, perhaps more correctly, exclude it. Their locus will be a pair of rays, yours will lack the end points.


I would not have done all this work, but just used the quadratic formula to find the x-coordinates of the two points of intersection, averaged them, and then put that into the equation of the line to find the y-coordinate of the midpoint of the chord.

What they have done is to find the x-coordinate using the Vieta trick, then find the average of the y-coordinates using another trick, and then eliminate n. The solution requires nothing more: you have the equation of the locus, and the restriction due to requiring a chord.


I don't see why you say there is still a need to find the y-coordinates of the end points of the chord, and thereby both coordinates of the end point of the locus (ray). The solution is, as you say, "y = 1/4 * x, FOR [imath]|x| >= 8\sqrt(3)/3[/imath]", which doesn't require that.

But, again, the more direct approach is, as you say, not to use all the trickery to find the averages, but just find the x-coordinates of the endpoints of the chord, average them, find y from the equation of the line, and then state the answer.

The "geometric placement of the middles of chords" is the locus, namely the equation satisfied by the midpoint of every such chord.


The restriction on n for forming a chord is needed to determine which part of the line y = x/4 forms the locus. That could instead have been done using the discriminant.

Here is a graph of the hyperbola (red), a chord (blue), its midpoint (green), and the locus (green):
View attachment 31650
The locus consists of all places the green point could be.

Does that make things any clearer?

You answered and clarified most of the questions I had about this problem ( the graph was particularly helpful ). Thank you for your help and I greatly applaud your enthusiasm.

However, I don't quite understand YOUR solution, which purportedly bypasses the ''tricks'' in the official solution. If I just do the quadratic and get the values of x, it will be dependent of n, right ( just the quadratic contains n ). How would you be able to get the x ( or y ) coordinate in the form it is in in the official ( a concrete number, a constant ) simply by applying the quadratic formula ?

Thank you once again for your aid.

 

[Dr.Peterson]

However, I don't quite understand YOUR solution, which purportedly bypasses the ''tricks'' in the official solution. If I just do the quadratic and get the values of x, it will be dependent of n, right ( just the quadratic contains n ). How would you be able to get the x ( or y ) coordinate in the form it is in in the official ( a concrete number, a constant ) simply by applying the quadratic formula ?

We have the equation [imath]3x^2+8nx+4n^2+16=0[/imath], whose solutions are [math]x=\frac{-4n\pm2\sqrt{n^2-12}}{3}[/math]
Taking the average, the sum of the solutions cancels the radicals, and dividing by 2 gives [math]x=\frac{-4n}{3}[/math] This is dependent on n; but we aren't finished. Now we put this value of x into the equation of the line, [imath]y=x+n[/imath], and get [math]y=\frac{-4n}{3}+n=\frac{-n}{3}[/math] So we now have equations for both x and y of the midpoint in terms of n.

Now we need to eliminate n between [imath]x=\frac{-4n}{3}[/imath] and [imath]y=\frac{-n}{3}[/imath]. Dividing the second by the first, we get [imath]\frac{y}{x}=\frac{1}{4}[/imath], resulting in the equation [imath]y=\frac{x}{4}[/imath].

This is all pretty routine. There was a very similar problem recently here (also from Serbian, I believe). There we didn't deal with the restriction, and the locus, being a horizontal line, had a different feel; but otherwise we used the same method.

 

[Ognjen]

We have the equation [imath]3x^2+8nx+4n^2+16=0[/imath], whose solutions are [math]x=\frac{-4n\pm2\sqrt{n^2-12}}{3}[/math]
Taking the average, the sum of the solutions cancels the radicals, and dividing by 2 gives [math]x=\frac{-4n}{3}[/math] This is dependent on n; but we aren't finished. Now we put this value of x into the equation of the line, [imath]y=x+n[/imath], and get [math]y=\frac{-4n}{3}+n=\frac{-n}{3}[/math] So we now have equations for both x and y of the midpoint in terms of n.

Now we need to eliminate n between [imath]x=\frac{-4n}{3}[/imath] and [imath]y=\frac{-n}{3}[/imath]. Dividing the second by the first, we get [imath]\frac{y}{x}=\frac{1}{4}[/imath], resulting in the equation [imath]y=\frac{x}{4}[/imath].

This is all pretty routine. There was a very similar problem recently here (also from Serbian, I believe). There we didn't deal with the restriction, and the locus, being a horizontal line, had a different feel; but otherwise we used the same method.

I still don't quite understand the role of n here. Shouldn't it be ascribed to every CHORD line instead of the locus itself ? ( the locus equation doesn't include n ( y = 1/4*x ), while the chord equation does ( y = x + n )).
However, if what I said is the case, then each chord line ( at given x ) would basically have an infinite number of midpoints ( as long as [math]2\sqrt(3)[/math] is within the boundaries of the hyperbola ), and ALL the midpoints drawn ( for each x ) wouldn't be collinear, and thus wouldn't make a line.

I would highly appreciate an explanation for this, if possible.

 

[Dr.Peterson]

I still don't quite understand the role of n here. Shouldn't it be ascribed to every CHORD line instead of the locus itself ? ( the locus equation doesn't include n ( y = 1/4*x ), while the chord equation does ( y = x + n )).
However, if what I said is the case, then each chord line ( at given x ) would basically have an infinite number of midpoints ( as long as [math]2\sqrt(3)[/math] is within the boundaries of the hyperbola ), and ALL the midpoints drawn ( for each x ) wouldn't be collinear, and thus wouldn't make a line.

I would highly appreciate an explanation for this, if possible.

I don't understand your thinking. How can a single chord have infinitely many midpoints?

Each value of n defines a line, which (if it passes through the hyperbola) defines a chord. We have found the two endpoints of that chord, and from them the midpoint of the chord. So we found that, for every n subject to an inequality, the point with [imath]x=\frac{-4n}{3}[/imath] and [imath]y=\frac{-n}{3}[/imath] is the midpoint of a chord. (That is, these two equations form a parametric equation of the locus we are looking for.)

To find the equation of the locus, we just need to eliminate the parameter n, which gave us the equation [imath]y=\frac{x}{4}[/imath]. This tells us how x and y are related for any point on the locus, without dependence on n.

 

[Ognjen]

I don't understand your thinking. How can a single chord have infinitely many midpoints?

Each value of n defines a line, which (if it passes through the hyperbola) defines a chord. We have found the two endpoints of that chord, and from them the midpoint of the chord. So we found that, for every n subject to an inequality, the point with [imath]x=\frac{-4n}{3}[/imath] and [imath]y=\frac{-n}{3}[/imath] is the midpoint of a chord. (That is, these two equations form a parametric equation of the locus we are looking for.)

To find the equation of the locus, we just need to eliminate the parameter n, which gave us the equation [imath]y=\frac{x}{4}[/imath]. This tells us how x and y are related for any point on the locus, without dependence on n.

But how is it "without dependence on n" when the solution is clearly restricted to a certain range ? It's not just a [math]y = x/4[/math] line, right ? Do we just have to "intuitively" conclude that the line cannot pass through areas not bounded by the hyperbola, and follow it with an again intuitive inference that "the point of intersection between the hyperbola and the locus line has to be the "limit on the outer side of the hyperbola", thus everything before it ( in case of the negative point ) and after it ( in the case of the positive point ) is included in the locus, and everything beyond that isn't ?

 

[Dr.Peterson]

But how is it "without dependence on n" when the solution is clearly restricted to a certain range ? It's not just a [math]y = x/4[/math] line, right ? Do we just have to "intuitively" conclude that the line cannot pass through areas not bounded by the hyperbola, and follow it with an again intuitive inference that "the point of intersection between the hyperbola and the locus line has to be the "limit on the outer side of the hyperbola", thus everything before it ( in case of the negative point ) and after it ( in the case of the positive point ) is included in the locus, and everything beyond that isn't ?

Of course it isn't just "intuitive". You've seen how it can be determined mathematically. You seem to keep changing which part of the problem you complain about.

The final answer doesn't contain n, but you use it in the process. In my work in #4, I didn't cover the part about the restriction, because I assumed you were satisfied with the inequality part. But we can add it:

We have the equation [imath]3x^2+8nx+4n^2+16=0[/imath], whose solutions are [math]x=\frac{-4n\pm2\sqrt{n^2-12}}{3}[/math]
Taking the average, the sum of the solutions cancels the radicals, and dividing by 2 gives [math]x=\frac{-4n}{3}[/math] This is dependent on n; but we aren't finished. Now we put this value of x into the equation of the line, [imath]y=x+n[/imath], and get [math]y=\frac{-4n}{3}+n=\frac{-n}{3}[/math] So we now have equations for both x and y of the midpoint in terms of n.

Now we need to eliminate n between [imath]x=\frac{-4n}{3}[/imath] and [imath]y=\frac{-n}{3}[/imath]. Dividing the second by the first, we get [imath]\frac{y}{x}=\frac{1}{4}[/imath], resulting in the equation [imath]y=\frac{x}{4}[/imath].

This is all pretty routine. There was a very similar problem recently here (also from Serbian, I believe). There we didn't deal with the restriction, and the locus, being a horizontal line, had a different feel; but otherwise we used the same method.

The quadratic equation has a solution only when the discriminant, [imath]n^2-12[/imath], is non-negative (and since when it is zero, there is no actual chord, so we can reject that case as well, though the author evidently didn't). This leads to the inequality [imath]n^2-12>0[/imath], so [imath]|n|>2\sqrt{3}[/imath]. This is a restriction on n. Since [imath]x=\frac{-4n}{3}[/imath], this implies that [imath]|x|>\frac{-4\cdot2\sqrt{3}}{3}=\frac{-8\sqrt{3}}{3}[/imath].

All of this is done mathematically, to conclude that the locus is the set of all points such that [imath]y=\frac{x}{4}[/imath], subject to the restriction that [imath]|x|>\frac{-8\sqrt{3}}{3}[/imath].

 

[Ognjen]

Of course it isn't just "intuitive". You've seen how it can be determined mathematically. You seem to keep changing which part of the problem you complain about.

The final answer doesn't contain n, but you use it in the process. In my work in #4, I didn't cover the part about the restriction, because I assumed you were satisfied with the inequality part. But we can add it:


The quadratic equation has a solution only when the discriminant, [imath]n^2-12[/imath], is non-negative (and since when it is zero, there is no actual chord, so we can reject that case as well, though the author evidently didn't). This leads to the inequality [imath]n^2-12>0[/imath], so [imath]|n|>2\sqrt{3}[/imath]. This is a restriction on n. Since [imath]x=\frac{-4n}{3}[/imath], this implies that [imath]|x|>\frac{-4\cdot2\sqrt{3}}{3}=\frac{-8\sqrt{3}}{3}[/imath].

All of this is done mathematically, to conclude that the locus is the set of all points such that [imath]y=\frac{x}{4}[/imath], subject to the restriction that [imath]|x|>\frac{-8\sqrt{3}}{3}[/imath].

Oh I get it now, the x restriction is dependent on n, so that's how n remains being relevant to the end solution. Thank you a lot and sorry about the inconvenience.
I kept changing the thing I complained about because of the new questions arising ( or resurging ) from your exposition; it may have confused you because I forgot to thank you and clarify that I do understand the thing you tried explaining formerly.

I hope I didn't annoy you.

Best wishes,
Ognjen