Equation of parabola passes through $4$ distinct points

Question

Equation of axis of parabola which

passes through the point $(0,1)\ , \ (0,2)$

And $(2,0)\ ,\ (2,2)$ is

Let general equation of conic is

$ax^2+2hxy+by^2+2gx+2fy+c=0\cdots (1)$

And it represent parabola if $h^2=ab$

Parabola passes through $(0,1)$

Then put into $(1)$

$b+2f+c=0\cdots(2)$

Also parabola passes through $(0,2)$

Then $4b+4f+c=0\cdots (3)$

Also parabola passes through $(2,0)$

Then $4a+4g+c=0\cdots (4)$

Also parabola passes through $(2,2)$

Then $4a+8h+4b+4g+4f+c\cdots (5)$

$(4a+4g+c)+(4b+4f+c)+8h-c=0$

From $(2)$ and $(3)$, we get $\displaystyle h=\frac {c}{8}$

From $(2)$ and $(3)$

$\displaystyle b+2f=4b+4f\Longrightarrow 2f=-3b $

Put into $(2)$ and $(3)$

$\displaystyle b=\frac{c}{2}$ and $\displaystyle f=-\frac{3c}{4}$

And $\displaystyle h^2=ab\Longrightarrow \frac{c^2}{64}=\frac{ac}{2}\Longrightarrow a=\frac{c}{32}$

Put all into $(4)$

$\displaystyle \frac{4c}{32}+4g+c=0\Longrightarrow g=-\frac{9c}{32}$

Put all values into $(1)$

$\displaystyle \frac{c}{32}x^2+\frac{c}{4}xy+\frac{c}{2}y^2-\frac{9}{16}x-\frac{3c}{2}y+c=0$

$x^2+8xy+16y^2-18x-48y+32=0$

Buti did not know how I find axis of parabola

Please have a look

Answer 1

What you've done is correct.

One way to find the axis is to write the equation as $(x+4y)^2-18x-48y+32=0,$ then you know the axis is parallel to $x+4y=0.$ The tangent at vertex is orthogonal and parallel to $4x-y.$ Now you can find where $4x-y+c$ intersects doubly to find the vertex. $$x^2+8x(4x+c)+16(4x+c)^2-18x-48(4x+c)+32=0$$ or $$289x^2+(136c-210)x+16c^2-48c+32=0$$ and this has discriminant $$(136c-210)^2-4\cdot 289\cdot (16c^2-4c+32)=(-4)(408c - 1777)$$ so the discriminant is zero for $c=\frac{1777}{408}.$ The vertex is the intersection of the tangent at vertex and the parabola. So solve $$x^2+8x(4x+\frac{1777}{408})+16(4x+\frac{1777}{408})^2-18x-48(4x+\frac{1777}{408})+32=0$$ or $$(1734x+1147)^2\frac1{10404}$$ and put back into the tangent at vertex to find $y.$ The vertex then is $(-\frac{1147}{1734},\frac{11857}{6936}).$ The axis goes through the vertex so the axis is $${ x+4y=\frac{105}{17}}.$$ As a bonus the tangent at vertex form is then $$(x+4y-105/17)^2=(96/(4\cdot 17))(4x-y+1777/408).$$ For fun I've also found the focus/directrix form $$17\cdot ((x+\frac{59}{102})^2+(y-\frac{689}{408})^2-(4x-y+\frac{113}{24})^2/17)=0$$

Answer 2

I'm going to show a solution which uses the following claim, and then add a proof of the claim and an important fact which might interest you.

Claim : The equation of a parabola can be written as $$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$ where $f(x,y)=0$ is the equation of the axis of symmetry, and $g(x,y)=0$ is the equation of the tangent at vertex. (Note that the axis of symmetry is perpendicular to the tangent at vertex.)

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Solution :

You correctly got $$x^2+8xy+16y^2-18x-48y+32=0$$ which can be written as $$(x+4y+c)^2+(-2c-18)x+(-8c-48)y-c^2+32=0$$

Since we want to find $c$ such that the line $x+4y+c=0$ is perpendicular to the line $(-2c-18)x+(-8c-48)y-c^2+32=0$, solving $$1\times (-2c-18)+4\times (-8c-48)=0$$ gives $c=-\dfrac{105}{17}$, and so the equation of the axis of symmetry is $x+4y-\dfrac{105}{17}=0$.

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In the following, I'll add a proof of the claim.

Claim : The equation of a parabola can be written as $$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$ where $f(x,y)=0$ is the equation of the axis of symmetry, and $g(x,y)=0$ is the equation of the tangent at vertex.

Proof :

The equation $$ax^2+2bxy+cy^2+2dx+2fy+g=0\tag2$$ represents a parabola iff $ac=b^2$ and $$\begin{vmatrix} a & b & d \\ b & c & f \\ d & f & g \\ \end{vmatrix}\not=0\tag3$$ (see here)

Multiplying the both sides of $(2)$ by $a$, and letting $$A=a,C=b,D=2ad,E=2af,F=ag$$ we see that, in general, the equation of a parabola is given by $$(Ax+Cy)^2+Dx+Ey+F=0\tag4$$ (where $(A,C)\not=(0,0)$, and $(3)\iff CD-AE\not=0$) which can be written as $$\bigg(f(x,y)\bigg)^2+g(x,y)=0$$ where $$\begin{align}f(x,y)&=Ax+Cy+\frac {AD+CE}{2(A^2+C^2)} \\\\g(x,y)&=\frac{CD-AE}{A^2+C^2}(Cx-Ay+G) \\\\G&=\frac{4F(A^2+C^2)^2-(AD+CE)^2}{4(A^2+C^2)(CD-AE)}\end{align}$$ Note that $f(x,y)=0$ is perpendicular to $g(x,y)=0$.

Now, we can see the followings by some calculations :

• The line $g(x,y)=0$ is tangent to the parabola $(4)$ at $P\bigg(H,\dfrac{CH+G}{A}\bigg)$ where $H=\dfrac{-2CG(A^2+C^2)-A(AD+EC)}{2(A^2+C^2)^2}$.

• The line $f(x,y)=0$ intersects the parabola $(4)$ only at $P$.

From these, we can say the followings :

• $f(x,y)=0$ is the equation of the axis of symmetry

• $g(x,y)=0$ is the equation of the tangent at the vertex

• $P$ is the vertex of the parabola.$\quad\blacksquare$

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Finally, I'll add an important fact which might interest you.

The equation of a parabola $$(Ax+Cy)^2+Dx+Ey+F=0$$ can be written as $$\bigg(\frac{f(x,y)}{\sqrt{A^2+C^2}}\bigg)^2=\frac{AE-CD}{(A^2+C^2)^{\frac 32}}\cdot\frac{Cx-Ay+G}{\sqrt{A^2+C^2}}$$

where

• $\dfrac{|f(x,y)|}{\sqrt{A^2+C^2}}$ represents the distance from $(x,y)$ to the axis of symmetry

• $\dfrac{|AE-CD|}{(A^2+C^2)^{\frac 32}}$ represents the length of its latus rectum (see here)

• $\dfrac{|Cx-Ay+G|}{\sqrt{A^2+C^2}}$ represents the distance from $(x,y)$ to the tangent at the vertex.

Answer 3

General equation is

$ A x^2 + B x y + C y^2 + D x + E y + F = 0 $

Plug in the points, you get the following four equations

$ C + E + F = 0 $

$ 4 C + 2 E + F = 0 $

$ 4 A + 2 D + F = 0 $

$ 4A + 4 B + 4 C + 2 D + 2 E + F = 0 $

This is a linear system of $4$ equations in $6$ unknowns. I fed this system of equations to a solver that I developed myself, and it gave me the following solution

$ (A, B, C, D, E, F) = t (-0.5, 0, 0, 1, 0, 0) + s (-0.25, 0.25, 0.5, 0, -1.5, 1 ) $

Setting $t = 2 \alpha$ and $ s = 4 \beta $, the solution can be written as

$ (A, B, C, D, E, F) = \alpha (-1, 0, 0, 2, 0, 0) + \beta (-1, 1, 2, 0, -3, 1) $

A necessary condition on these parameters to have a parabola is that $B^2 - 4 A C = 0$. If $\alpha = 0 $, then $B^2 - 4 A C = \beta^2 (1 + 8) = 9 \beta^2 \ne 0 $. Therefore, we can assume that $\alpha \ne 0 $ and pull it out and divide by it, to get

$ (A, B, C, D, E, F) = (-1, 0, 0, 2, 0, 0) + \lambda (-1, 1, 2, 0, -3, 1) $

where $\lambda = \beta / \alpha $

Now the parabola condition becomes,

$ \lambda^2 - 4 (-1 - \lambda)(2 \lambda) = 0 $

Which give two values for $\lambda$, namely $\lambda = 0 $ and $\lambda = - 8/ 9 $

If $\lambda = 0 $ , then we have

$ - x^2 + 2 x = 0 $

which is not a parabola, but a pair of parallel lines. With the other solution

$ - \dfrac{1}{9} x^2 - \dfrac{8}{9} xy - \dfrac{16}{9} y^2 + 2 x + \dfrac{8}{3} y - \dfrac{8}{9} = 0 $

Mutliplying through by $(-9)$

$ x^2 + 8 x y + 16 y^2 - 18 x - 24 y + 8 = 0 $

is the only possible parabola.