HSC Stuff > HSC Mathematics Advanced

Locus and the parabola question

(1/1)

_OwO_:
Hey! please send help
how would you do this question:

find the equation of the parabola with the coordinates of the vertex (0,0)  and equation of the axis  x=0  and passing through the point (-8,2)

thankyou!

jamonwindeyer:
Hey there! Super sorry for the late reply my friend, sort of stopped looking at the maths boards for a bit in the HSC rush - Keen to give you a hand though ;D

So the first two bits of info are key here. We have a parabola with a vertex at \((0,0)\) and an axis of \(x=0\) (the y-axis). This means we have a parabola that is vertical, with its turning point at the origin (just a really plain, vanilla parabola really!). Any parabola on the origin like this takes the form:



To find the actual equation, we need \(a\). To get it, substitute the point we are given:



So substituting that in, we get \(x^2=32y\), or \(y=\frac{x^2}{32}\). That is the equation we need! I hope this helps ;D

Navigation

[0] Message Index