GCSE Maths Quadratic Simultaneous Equation Loci Watch

theteknikal
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I know someone already posted this but I struggled to understand what they were trying to convert. Anyway much help would be appreciated solving this problem:
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The diagram shows a sketch of a curve. The point P(x,y) lies on the curve [i.e the curve is the locus of the point P.]
The locus of P has the following property: the distance of the point P from point (0,2) is the same as the distance from the x-axis.
Show that y=(1/4)x^2 +1.
"
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BobBobson
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So you need to model the distance point P is from (0,2) algebraically. I assume you know how to find the distance of a point (using pythagoras' thereom).

You also know that the horizontal length of the imaginary triangle is x. And you know that the vertical length is y-2 and you need to find the hypotenuse (which will be the distance).

Another way of writing the distance from the x axis is just y. So y = hypotenuse of the triangle.

You just rearrange the equation a bit, complete the square, rearrange some more, bing bang boom. And you get the answer.
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Gingerbread101
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(Original post by theteknikal)
I know someone already posted this but I struggled to understand what they were trying to convert. Anyway much help would be appreciated solving this problem:
Name:  2016-11-10 06.37.49 1.jpg
Views: 112
Size:  497.6 KB
The diagram shows a sketch of a curve. The point P(x,y) lies on the curve [i.e the curve is the locus of the point P.]
The locus of P has the following property: the distance of the point P from point (0,2) is the same as the distance from the x-axis.
Show that y=(1/4)x^2 +1.
"
I've moved this thread to the Maths forum, you'll get more replies here as the GSCE forum isn't for subject specific questions
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RDKGames
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(Original post by theteknikal)
I know someone already posted this but I struggled to understand what they were trying to convert. Anyway much help would be appreciated solving this problem:

The diagram shows a sketch of a curve. The point P(x,y) lies on the curve [i.e the curve is the locus of the point P.]
The locus of P has the following property: the distance of the point P from point (0,2) is the same as the distance from the x-axis.
Show that y=(1/4)x^2 +1.
"
Right so you're given information which you should follow. You know P(x,y) and you can leave it's coordinates in that form. The point (0,2) is what we call the focus of the parabola and the equation of it stems from this particular point. A focus of a parabola is simply the point for which the distance from it to some point is the same as the distance from the x-axis, and if you were to sketch a lot of these points down, you'd get a parabola all the time.

First find the distance from the x-axis to point P (which should be extremely obvious)

Secondly, find the distance from (0,2) to P which involves using Pythagoras' Theorem, and I assume you know how to find the distance between 2 points.

Thirdly, equate your two answers since you know these two distances are the same, then just rearrange for y.
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