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Was wondering why $y^2 =4ax$ is considered as the standard form of a parabola? I was going through my textbook and all important results like equation of Tangent, Normal, Chord of contact was given in terms of $y^2 = 4ax$ form. So is there any way to convert between them?

Alberto Debernardi
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Mahadev
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  • I believe that traditional form is used because then $a$ is the distance between the vertex and the focus. – Leonard Blackburn Dec 04 '17 at 15:29
  • Probably because the standard formula is in fact a special case of general formula for parabola, ellipse and hyperbola: $y^2=2px-(1- \epsilon^2)x^2$ in which $\epsilon =0$ and $p =2a$ for parabola. – sirous Dec 04 '17 at 15:36
  • @Mahadev If you are ok, you can set as solved. Thanks! – user Dec 06 '17 at 13:14

2 Answers2

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You can convert any parabola into any other parabola by completing the squares and shifting around. For instance, you are given

$$x=Ay^2+By+C$$ You first complete the square: $$x=A(y^2+(B/A)y+C/A)=A(y+B/(2A))^2+C/A-B^2/(4A)^2$$ Rewrite

$$(y-y_0)^2=A^{-1}(x-x_0)$$ where $y_0=-B/(2A)$ is how much the parabola is moved in $y$ direction from the standard form, $x_0=C/A-B^2/(4A)^2$ is how much it is moved in $x$ direction and $A^{-1}$ is the equivalent of your $4a$ parameter.

If your parabola has $x$ in squares and $y$ in linear terms, just exchange $x$ and $y$.

orion
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By switching the axes: $$y^2 =4ax$$ is totally equivalent to $$x^2 =4ay \implies y=\frac{x^2} {4a}$$.

user
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  • I am given a formula for equation of tangent at a specific point (x1,y1) on parabola y^2=4ax as yy1=2a(x+x1) how would I write a similar equation for x^2=4ay? – Mahadev Dec 04 '17 at 15:45
  • For $y^2=4ax$ the tangent at $(x_1,y_1)$ should be $y-y_1=2a(x-x_1)$. – user Dec 04 '17 at 15:50
  • The most widely used form is $y=4ax^2$ with x as horizontal axis and y vertical. In this case the parabola is concave up. – user Dec 04 '17 at 15:57