# A doubt while solving a question on parabola.

Question:

Find the equation of the parabola whose latus rectum is $$4$$ units,axis is the line $$3x+4y-4=0$$ and the tangent at the vertex is the line $$4x-3y+7=0$$.

This is the solution

I have marked the part(in the image) which is troubling me.

The problem part is $$PM²=(\text{latus rectum}) ×PN$$.

I read on a site that it is the general equation of parabola, i.e.,$$y²=4ax$$,that is used here, but in the general equation, $$y$$ and $$x$$(Co-ordinates of the point $$P$$).In my problem part they are replaced by $$PM$$ and $$PN$$ respectively,both of which are surely not equal($$PM≠y$$ and $$PN≠x$$) .see this image, it will be more clear.

Also, since the axis of parabola is not parallel to any of the axes, so the equation of the parabola must contain a $$xy$$ term with non-zero co-efficient. But if we write $$y²=4ax$$, that co-efficient becomes Zero. So I think that $$PM²=(\text{latus rectum})×PN$$ is not used as the equation of parabola but as a condition.

And if it is a condition, how we came to that. What's the logic/mathematics. I think that if we rotate and move the parabola then $$PM=y$$ and $$PN=x$$.

Please clear this to me. I am really having a hard time with this.

We know that the latus rectum, which I will denote by $$LR$$ for short, is the line segment through the focus whose endpoints are on the parabola and perpendicular to the parabola's axis. The idea then, is that the length of $$LR$$ is related to the distance between the focus and vertex.

When we look at the equation of a parabola in its simplest form, say $$y = f(x)= \frac{1}{4a} x^2$$ for some $$a > 0$$, where is the focus? It is the point $$(0,c)$$ that satisfies $$(x-0)^2 + \left(\frac{x^2}{4a} - c\right)^2 = (x-x)^2 + \left(\frac{x^2}{4a} + c\right)^2,$$ because the locus of a parabola is the set of points equidistant from the focus and the directrix. Solving this for $$c$$ yields $$c = a,$$ which explains our choice of parametrization of the coefficient in $$f(x)$$. Thus $$a$$ represents the distance between the vertex and focus.

The length of $$LR$$, then, is simply twice the magnitude of the $$x$$-value for which the $$y$$-value equals $$a$$; i.e., it is $$|LR| = 2|2a| = 4|a| = 4a,$$ since $$a > 0$$ was assumed. But if this is true, then $$4ay = x^2$$ becomes $$|LR| y = x^2.$$ That also means that no matter what isometry is applied to the coordinate system, as long as the axes are perpendicular, $$|LR||PN| = |PM|^2$$ where $$|PN|$$ is the distance to the tangent axis (the $$y$$-coordinate) and $$|PM|$$ is the distance to the axis (the $$x$$-coordinate).

• seems that you have not got my problem. I know these derivations and all. May 15 '20 at 16:45
• "PM≠y and PN≠x" is the problem May 15 '20 at 16:46
• @Kartikey As I have stated: "no matter what isometry is applied to the coordinate system..." what this means is that any rotation/translation/reflection of the the coordinate system that maps the axes of the parabola to the coordinate axes, the underlying geometric relationship holds. May 15 '20 at 16:47
• I should point out that when I went through this derivation, I did not state that the coordinate system I used was the coordinate system of the rotated/translated parabola in your problem. Instead, it is the coordinate system of the lines corresponding to the parabola's axis and the vertex tangent. May 15 '20 at 16:50
• So if we rotate and move the axes, then the equation remains as it Is. no xy term gets introduced and the coefficients remains the same. May 15 '20 at 16:54

The condition $$PM^2=(latus\, rectum)\times PN$$ is a geometric condition defining the parabola. You replace $$PM$$ and $$PN$$ by the formulas giving you the distances from a point to a straight line. That is all you do.

Whatever coordinates you use, if you compute $$PM$$ and $$PN$$ in those coordinates and set $$PM^2=(latus\, rectum)\times PN$$ you get the equation of the parabola in those coordinates.

Precisely because in your case $$PM\neq y$$ and $$PN\neq x$$ you do not get the canonical form (your axes are not optimally chosen)

$$PM^2=(\text{latus rectum})\times PN$$ is the equation of the parabola. This follows from the fact that if you choose the axes in the standard position ($$x$$-axis is the axis of the parabola and $$y$$-axis is tangent at the vertex, then the equation of the parabola is $$y^2=4px$$, and $$4p$$ is the length of the latus rectum. The equation of the final parabola contains $$xy$$ terms, as you can verify expanding the square on the left hand side.

• if we choose the axes in standard position the equations of axis and the tangent too changes. It does not seems to get reflected over there (in rest part of the solution) May 15 '20 at 16:43

Okay so you have a doubt in how $$PM=y$$ and $$PN=x$$. When we write the equation $$y^2=4ax$$, the $$PM$$ stands for axis of the parabola and $$PN$$ stands for the tangent at the vertex.

For the parabola $$y^2=4ax$$ the $$x$$-axis i.e. $$y=0$$ is the axis and the $$y$$-axis i.e. $$x=0$$ is the tangent at the vertex. Hence it becomes $$y^2=4ax$$.

For your given axis and tangent at the vertex, equation of the parabola will be (axis)$$^2=$$(latusrectum)*(tangent at vertex).

That is,

$$(3x+4y-4)^2 = 4(4x-3y+7)$$ which does have a $$xy$$ ter with non zero coefficient.

It simplifies to $$9x^2+16y^+24xy-40x-20y-12=0$$. Please comment if this is what you wanted to ask.

I think the piece that’s missing for you is a different interpretation of Cartesian coordinates than you may be used to. You’re probably used to thinking of the standard Cartesian coordinates $$(x,y)$$ in terms of a grid of some sort, or moving along the axes. Instead of that, think of the $$x$$-coordinate of a point as measuring the (signed) distance from the $$y$$-axis, i.e., from the line $$x=0$$. Similarly, the $$y$$-coordinate measures the distance of the point from the $$x$$-axis, namely the line $$y=0$$. Observe that if you use the standard formula of a point from the line $$x=0$$, you just get $$x$$.

With this interpretation (and ignoring the effect of signs, which is easy to accommodate) what the standard parabola equation $$y^2=4px$$ says is that the parabola is the set of points for which the square of the distance to the $$x$$-axis—i.e., the square of the distance $$PM$$ to the parabola’s axis—is equal to the latus rectum $$4p$$ times the distance to the $$y$$-axis—the distance $$PN$$ to the tangent at the vertex. This is a coordinate-free characterization of a parabola: it holds regardless of which pair of perpendicular lines you choose for the axis and vertex tangent.

What does this equation look like with two arbitrary perpendicular lines, then? Well, we can always arrange things so that the equation of the axis has the form $$ax+by+c=0$$ and the equation of the perpendicular is $$bx-ay+d=0$$. Now, just use the standard point-line distance formula: the distance to the parabola’s axis, $$PM={\lvert ax+by+c\rvert\over\sqrt{a^2+b^2}}$$ and the distance to the vertex tangent is $$PN={\lvert bx-ay+d\rvert\over\sqrt{a^2+b^2}}$$. Points on the parabola therefore satisfy $${(ax+by+c)^2\over a^2+b^2}=4p{\lvert bx-ay+d\rvert\over\sqrt{a^2+b^2}},$$ which is what was used in the solution to the problem.