weird conic section problem: Draw y^2=x+1, 2 tangents passing thru (-3,0)

[ka923]

A parabola P has equation y² = x + 1

(a) Sketch P and, on your sketch, draw two tangents to P passing through the point (-3, 0).

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for this question do i just draw 2 random tangent that crosses the point?
btw i have already sketched p

 

[Dr.Peterson]

A parabola P has equation y² = x + 1

(a) Sketch P and, on your sketch, draw two tangents to P passing through the point (-3, 0).

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for this question do i just draw 2 random tangent that crosses the point?
btw i have already sketched p

They won't be "random" tangents; they will be the two tangents that pass through that point. Presumably you are expected to locate them accurately, though it's also possible that you are just supposed to sketch the whole figure roughly, for some reason.

The context can be very important.

What do you know about tangents to parabolas? You didn't put this under calculus; do you know about derivatives, so that you could write an equation to find the point of tangency? Or have you learned something geometrical about them?

 

[ka923]

um i dont think so as we have just begun to learn about parabolas

 

[Dr.Peterson]

um i dont think so as we have just begun to learn about parabolas

Then I guess they just want you to draw a picture, as accurately as you want.

Can you tell us anything more? See our guidelines for suggestions. Clearly you have quoted the problem completely; but has anything else been said about tangents at all?

 

[ka923]

A parabola P has equation y² = x + 1

(a) Sketch P and, on your sketch, draw two tangents to P passing through the point (-3, 0).

(b) Show that, if the line y = m (x + 3) intersects P, then the x-coordinates of the points of intersection must satisfy the following equation:

. . . . .m² x² + (6m² - 1) x + (9m² - 1) = 0

(c) Show that, if y = m (x + 3) is a tangent to P, then m² = 1/8.

(d) Hence, find the exact coordinates of the points at which these tangents touch P.

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this is the entire set of the question, i guess it has something to do with d

 

[Dr.Peterson]

A parabola P has equation y² = x + 1

(a) Sketch P and, on your sketch, draw two tangents to P passing through the point (-3, 0).

(b) Show that, if the line y = m (x + 3) intersects P, then the x-coordinates of the points of intersection must satisfy the following equation:

. . . . .m² x² + (6m² - 1) x + (9m² - 1) = 0

(c) Show that, if y = m (x + 3) is a tangent to P, then m² = 1/8.

(d) Hence, find the exact coordinates of the points at which these tangents touch P.

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this is the entire set of the question, i guess it has something to do with d

Okay, it looks like they just want an approximate sketch first, so you see what the situation is like. Then they are leading you through a non-calculus way to identify the exact equation of a tangent line, by solving for the slope at which the line will intersect only once. Then you can check how close your sketch was.

 

[HallsofIvy]

A parabola P has equation y² = x + 1

(a) Sketch P and, on your sketch, draw two tangents to P passing through the point (-3, 0).

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for this question do i just draw 2 random tangent that crosses the point?
btw i have already sketched p

No, not "random" tangents! There are only two lines through (-3, 0) that are tangent to $\displaystyle y^2= x+ 1$. Before you can draw them you have to determine what they are! Any line can be written y= mx+ b[/tex]. The fact that the line goes through (-3, 0) means that 0= -3m+ b so that b= 3m and we can write the equation as y= mx+ 3m. That line must "touch" the parabola at some point $\displaystyle (x_0, y_0)$ so $\displaystyle (x_0, y_0)$ must satisfy both equations: $\displaystyle y_0= mx_0+ 3m$ and $\displaystyle y_0^2= x_0+ 1$. That gives us two equations but to find the three unknowns, $\displaystyle x_0$, $\displaystyle y_0$, and m, we need a third equation.

That is given by the condition that the line is tangent to the parabola: the slope of the line, m, is equal to the dy/dx for the parabolic equation. $\displaystyle y^2= x+ 1$ so, differentiating both sides with respect to x, $\displaystyle 2y (dy/dx)= 1$ and, at $\displaystyle (x_0, y_0)$ the derivative is $\displaystyle dy/dx= \frac{1}{2y_0}$.

You need to solve the equations $\displaystyle y_0= mx_0+ 3m$, $\displaystyle y_0^2= x_0+ 1$ and $\displaystyle m= \frac{1}{2y_0}$ for $\displaystyle x_0$, $\displaystyle y_0$, and m. Two of those equations are quadratic so there will be two solutions for m. They will give you the two tangent lines.

 

[JeffM]

My problem with this problem is that we do not know what an algebra student is expected to know.

No, not "random" tangents! There are only two lines through (-3, 0) that are tangent to $\displaystyle y^2= x+ 1$.

It is not trivial to prove that, if a tangent to a given parabola runs through a given point not on the parabola, at most one other tangent to the same parabola runs through the same point. The easiest way to prove it is by calculus, but the easiest way to discover it is with a sketch.

Before you can draw them you have to determine what they are!

This is not true of a sketch. If the student tried to sketch it, he would soon discover that, of the infinite number of lines emanating from the given point, two seemed to be tangents. And you can make a sketch without knowing the exact co-ordinates of the point of tangency.

Any line can be written y= mx+ b[/tex]. The fact that the line goes through (-3, 0) means that 0= -3m+ b so that b= 3m and we can write the equation as y= mx+ 3m. That line must "touch" the parabola at some point $\displaystyle (x_0, y_0)$ so $\displaystyle (x_0, y_0)$ must satisfy both equations: $\displaystyle y_0= mx_0+ 3m$ and $\displaystyle y_0^2= x_0+ 1$. That gives us two equations but to find the three unknowns, $\displaystyle x_0$, $\displaystyle y_0$, and m, we need a third equation.

That is given by the condition that the line is tangent to the parabola: the slope of the line, m, is equal to the dy/dx for the parabolic equation. $\displaystyle y^2= x+ 1$ so, differentiating both sides with respect to x, $\displaystyle 2y (dy/dx)= 1$ and, at $\displaystyle (x_0, y_0)$ the derivative is $\displaystyle dy/dx= \frac{1}{2y_0}$.

An algebra student should be fine with this up to the point where the derivative is brought into play. It is not so obvious how to proceed if you have no clue what differentiating is.

If we read the progression of the problem as given, it seems that it is designed to lead the student to discover that there are two such tangents and provide a purely algebraic way to find their equations, but I cannot be sure that is what is going on. We know nothing about the context in which this problem was posed.

My questions to the OP: do you know differential calculus, and, if not, how far did you get with each part using just geometry and algebra?

 

[ka923]

hi sorry for the late reply, so far we have learned about the equation of the parabola and how to sketch it, also how to rearrange the equation of the parabola. we have not learnt about differentiation yet

 

[JeffM]

hi sorry for the late reply, so far we have learned about the equation of the parabola and how to sketch it, also how to rearrange the equation of the parabola. we have not learnt about differentiation yet

That is very helpful to know where you are in your studies. Differentiation is the topic of differential calculus so answers involving differentiation will not help you yet.

So, getting back to your problem, if you do the sketch requested in part a, it will indicate that there are at least two different lines that are tangent to the given parabola and that run through the given point. That is not a proof that there are 2, let alone that there are only 2, but it is designed to make you comfortable with the basis for the next questions.

Now in part b, you are asked about a generalized line y = m(x + 3). What the question does not say, but leaves for you to figure out, is that

$\displaystyle x = -\ 3 \implies y = m(-\ 3 + 3) = 0.$

In other words, the equation given to you describes a generalized line that runs through (- 3, 0) so it is not any old line. Furthermore, the question further restricts the line to those that intersect the parabola. Now it asks you to confirm something about that intersection. What we know about the intersection is that

\(\displaystyle y^2 = x + 1 \text { and } y = m(x + 3) \implies y^2 = m^2x^2 + 6m^2x + 9m^2 \implies\\

x + 1 = m^2x^2 + 6m^2x + 9m^2 \implies m^2x^2 + (6m^2 - 1)x + (9m^2 - 1) = 0.\)

That is all there is mechanically to do in b. Easy, peasy.

But notice that we have a quadratic equation. And we know that a quadratic equation can have 2 real roots, 1 real root, or no real roots, depending on whether the discriminant is positive, zero, or negative. If the discriminant is positive, the line intersects the parabola at TWO points. If the discriminant is zero, the line intersects the parabola at EXACTLY ONE point. And if the discriminant is negative, the line does not intersect the parabola at all.

Thus, we now have the general equation for the x co-ordinates of the intersection of the parabola and a line through (- 3, 0). But we started off asking about tangent lines.

Part c gets us back to tangents. It asks us to confirm something about lines that are tangent to the parabola and run through (- 3, 0). What do we know about tangents to a parabola? They intersect the parabola at EXACTLY ONE point. So the discriminant to the equation that we found in part b must be 0.

What is that discriminant?

$\displaystyle (6m^2 - 1)^2 - 4(m^2)(9m^2 - 1) = 36m^4 - 12m^2 + 1 - 36m^4 + 4m^2 = 1 - 8m^2$

And if that discriminant is zero, then

$\displaystyle 0 = 1 - 8m^2 \implies 8m^2 = 1 \implies m^2 = \dfrac{1}{8}.$

Again, the mechanics are easy. But what are the implications? We have ANOTHER quadratic, in m this time, with two real solutions. Thus, we have confirmed that there are exactly two lines that run through the given point and are tangent to the given parabola. Moreover, if we solve for m, we can go back to b to get the x co-ordinates where the two lines are tangent. And then we plug into the equation for the parabola to get the y co-ordinates. And that is what d is asking us to do. So off you go.

I hate this problem because it asks you to do purely mechanical exercises without telling you why you are doing them. Steps b and c just ask you to check the author's work in getting to step d, but they don't explain why the author did those steps.

I hope this helps to dispel what is completely unnecessary fog.