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Given a line $y=kx$ on a Cartesian coordinate, I want to find an equation of a parabola, whose base is on that line at point $(x_1,y_1)$ and passes through point $(x_2,y_2)$.

Then, an equation of a red line that is perpendicular to line $y=kx$ at point $(x_1,y_1)$ is $y=-\frac{1}{k}x+y_1+\frac{1}{k}x_1$. I tried to assume that $y=kx$ line is my new $x$ axis, and $y=-\frac{1}{k}x+y_1+\frac{1}{k}x_1$ is my new $y$ axis and do calculations, but the computation became messy and I couldn't finish. Is there simple way to solve this problem?

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insipidintegrator
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Lee
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2 Answers2

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From this post, we can write the equation of a parabola as $$d_{\text{Axis}}^2=4a\cdot d_{\text{Tangent at vertex}}.$$ Here $d_{\text{line}}$ represents the algebraic distance of any arbitrary point on the parabola from the said line. $**$(More intuitive explanation below).

Here, tangent at vertex is $y=kx$.
The axis (straight line through the vertex (that you referred to as the base) and perpendicular to tangent at vertex) is $y+\dfrac xk=y_1+\dfrac{x_1}{k}$.

Thus the equation of parabola is $$\bigg(\frac{ky+x-(ky_1+x_1)}{\sqrt{1+k^2}}\bigg)^2=4a\cdot\dfrac{y-kx}{\sqrt{1+k^2}}$$

Put $(x_2, y_2)$ into the equation and find $a$. Done :)

$**$ Intuitive explanation: Consider the standard parabola $y^2=4ax$. Here $y$= Distance from X axis ($\equiv$ Axis) and $x$= Distance from Y axis ($\equiv $ Tangent at vertex)

insipidintegrator
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$x_1$ and $y_1$ make the problem complicate. First, we do a translation to make $x_1, y_1$ move to origin in $x'-y'$ coordinate system.

Now, the parabola touch $y=kx$ at origin and pass $(x'_0,y'_0)$ where $x'_0=x_2-x_1, y'_0=y_2-y_1$.

Then we make a rotation to $x''-y''$ coordinate system. by a rotation of $\theta$.

$x'=x''\cos\theta + y''\sin\theta$

$y'=-x''\sin\theta + y''\cos\theta$ (Hope I did not rotation at wrong direction).

$(x'_0,y'_0)$ now rotated to $(x''_0,y''_0)$.

Then the parabola touch x-axis at origin.

The parabola is $y''=ax''^2$. Put $(x''_0,y''_0)$ into the equation, you can solve for $a$ and fix the parameter of parabola equation in $x''-y''$ coordinate. Then you can work back to a equation in $x-y$ coordinate. The equation could be complicated so you can write it in terms of $a$ since you already have the expression of $a$.

Method II: Start from general equation of parabola: $\frac{(ax+by+c)^2}{a^2+b^2}=(x-x_f)^2+(y-y_f)^2$

Assume the signed distance between focus $(x_f,y_f)$ to $y=kx$ is $d$. Then you can write down the equation of directrix $ax+by+c$ and focus $(x_f,y_f)$ in term of $d$. Then put $(x_2,y_2)$ into the equation, you can find $d$ and then fix all parameters $a,b,c,x_f,y_f$.

Abel Wong
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