Question involving calculations with a parabola

[Kaushik]

Summary: Find the equation of the parabola when the focus and the equation of tangent at the vertex is known.

Find the equation of the parabola when the focus is $(0,0)$ and the equation of tangent at the vertex is $x - y + 1 = 0$.

General equation : $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$
Apart from this I have another question. It is given in my textbook that $h^2 - ab = 0$ for the general equation to represent parabola. Why or from where is this derived?

 

[mfb]

This fits better into our homework section.

Can you express the focus as function of a,b,c,h,g,f? Similarly, the tangent at the vertex will lead to additional equation that you can use to find the coefficients.

It is given in my textbook that $h^2 - ab = 0$ for the general equation to represent parabola. Why or from where is this derived?

This is missing context what h, a and b are.

 

[Kaushik]

This is missing context what h, a and b are.

I think before going to HW question it would be better if you clarify this.
As I said $h, b$ and $a$ are coefficients from the general equation:
$ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$

 

[haruspex]

It is given in my textbook that $h^2 - ab = 0$ for the general equation to represent parabola. Why or from where is this derived

If you substitute that in the quadratic, what factorisation can you do?

 

[Kaushik]

If you substitute that in the quadratic, what factorisation can you do?

Substitute what?

 

[LCKurtz]

Another (different) approach. If you plot what you are given, you can easily see the vertex is at $(-\frac 1 2, \frac 1 2)$ and the directrix is $y = x + 2$. The definition of a parabola requires the distance from the parabola to the focus = the distance from the parabola to the directrix. You have some 45-45-90 right triangles to make the distances easy to calculate.

 

[Mark44]

Find the equation of the parabola when the focus is $(0,0)$ and the equation of tangent at the vertex is $x - y + 1 = 0$.

General equation : $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$
Apart from this I have another question. It is given in my textbook that $h^2 - ab = 0$ for the general equation to represent parabola. Why or from where is this derived?

The general equation you show above is not that of a parabola -- it is the general equation of a conic section, which could be a circle, parabola, ellipse, or hyperbola.
All of these conic sections have an $x^2$ term or a $y^2$ term (or both) as well as terms in x and y, and a constant term. If the conic section has been rotated, it will have an $xy$ term.

If you substitute that in the quadratic, what factorisation can you do?

I don't think that substituting $h^2 = ab$ will do anything here. This is just the relationship that parabolas have with regard to the general equation of a conic section.

#Kaushik, since you are given that the line $x - y + 1 = 0$ is tangent to the parabola at its vertex, it follows that the axis of symmetry of the parabola makes an angle of 45° with the y-axis, and this axis of symmetry runs through the 2nd and 4th quadrants.

I would start by drawing a sketch of this tangent line, then drawing a rough sketch of the parabola, which must open down into the 4th quadrant. The vertex of the parabola is on the line $x - y + 1 = 0$, and the focus is at (0, 0), so you should be able to find the coordinates of the vertex, and from that, p, which plays a role in the definition of a parabola.

This PDF, http://www.cengage.com/resource_uploads/downloads/1285060288_382318.pdf, might be helpful.

 

[Kaushik]

Why or from where is this derived?

@Mark44 Could you answer the above question? Or did I miss it in your previous post?

 

[Mark44]

I said this in my previous post.

All of these conic sections have an $x^2$ term or a $y^2$ term (or both) as well as terms in x and y, and a constant term. If the conic section has been rotated, it will have an $xy$ term.

 

[Kaushik]

I said this in my previous post.

But I don't find how that is related to $h^2 = ab$ condition for parabola.

 

[Mark44]

But I don't find how that is related to $h^2 = ab$ condition for parabola.

Take a look at the PDF I linked to earlier, especially Theorem D.1.

 

[haruspex]

Substitute what?

Use h²=ab to replace h in the quadratic.
It will be clearer what to do next if you replace a and b with their square roots: A²=a, B²=b, h=AB.

 

[Kaushik]

Use h²=ab to replace h in the quadratic.
It will be clearer what to do next if you replace a and b with their square roots: A²=a, B²=b, h=AB.

I tried substituting but It doesn't seem to help me either. I think I missing something here. I'm sorry but would be nice if you give few more hints?

 

[haruspex]

I tried substituting but It doesn't seem to help me either. I think I missing something here. I'm sorry but would be nice if you give few more hints?

After the substitution, what do the three quadratic terms look like? Can you see a factorisation? If you don't, post what you get.

 

[RPinPA]

I was thinking of a different approach, which I see is what @Mark44 is suggesting.

Let's start with the equation for a parabola which opens up along the $\pm y$ axis. This would have the form $y = r(x - s)^2 + t$ so it will have only $y$, $x$, $x^2$ and constant terms. Using your form, that means it has the form $ax^2+2gx+2fy+c=0$.

Then apply Theorem D.1 from @Mark44's reference, rotating the coordinate axes by an arbitrary $\theta$. That is, make the substitution $x = x' \cos \theta - y' \sin \theta$, $y = x' \sin \theta + y' \cos \theta$. This will give the equation for a parabola at an arbitrary angle relative to the $x,y$ axes.

If you get terms in $(x')^2$, $(y')^2$ and $x'y'$, see how the coefficients are related.

 

[haruspex]

Since this thread seems to have fizzled out, I will explain my approach in post #12.

After the substitution we get (Ax+By)²+2gx+2fy+c=0.
It follows that $2gx+2fy+c\leq 0$, so the curve is restricted to a half plane. Thus it is a parabola. Moreover, $2gx+2fy+c=0$ must be a tangent, though not necessarily at the vertex, and at the point of contact Ax+By=0.

What might not be quite so obvious is that the line Ax+By=0 must be parallel to the axis. In the present case, we are told the focus is at the origin. It follows that the axis goes through the origin, so Ax+By=0 is the axis and the tangent we found touches at the vertex.

Plugging in the given equation for that tangent we can start determining the coefficients.

 

[LCKurtz]

Can't leave the easiest method to die on the vine here . From the given we know the focus is at $(0,0)$ and the vertex is at $(-\frac 1 2,\frac 1 2)$. It follows that the directrix is $y = x+2$ from the geometry. Here's a picture with a couple of additions I will explain below:

If $P(x_p,y_p)$ is a point on the parabola, its distances $d$ from the directrix and focus are the same. Distance from the focus is $d=\sqrt{x_p^2+y_p^2}$. Distance from the directrix is the leg of the $45^\circ 45^\circ 90^\circ$ triangle $PQR$ with hypotenuse $h = x_p - (y_p-2)$ (the blue line). So $d =\frac{x_p-y_p+2}{\sqrt 2}$. Square these two expressions for $d$, set them equal, and drop the $p$ subscripts, and you have your equation.

 

[Kaushik]

vertex is at $(−1/2,1/2)$

How did you conclude that the vertex is at that point?

 

[mfb]

That's the point of the tangent of the vertex closest to the focus.

 

[Kaushik]

That's the point of the tangent of the vertex closest to the focus.

I'm not sure I understand

 

[LCKurtz]

I'm not sure I understand

Draw any parabola on a piece of paper and the tangent line at its vertex. The focus is on the axis of the parabola and the closest point on the tangent line to the focus is the vertex. Also the axis of the parabola and the tangent line are perpendicular at that point.

 

[Mark44]

Can't leave the easiest method to die on the vine here

This is exactly the approach I was suggesting in post #7.