Parabola question (1 Viewer)

V_L · Jan 17, 2017

Hi,

Can someone please go through how to do the following question? the answer is attached.

Find the equation of the parabola with coordinates of the vertex (0, 0) and equation of the axis y = 0 and passing through the point (3, 6)

Thank you

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1729 · Jan 17, 2017

V_L said:
Hi,

Can someone please go through how to do the following question? the answer is attached.

Find the equation of the parabola with coordinates of the vertex (0, 0) and equation of the axis y = 0 and passing through the point (3, 6)

Thank you )

The answer you attached does not have axis y = 0 nor does the point (3,6) belong to the curve.

$$\noindent We know that the curve is of the form $y^{2}=4ax.$ Note that I did not write $\pm 4a$, as a parabola with vertex the origin cannot pass through a point in the first quadrant if it's concavity is leftwards, which occurs for $y^{2} = -4ax \\ \\ $Now, we substitute the coordinate (3,6) into $y^{2} = 4ax$ and obtain $a = 3$. Thus, such a parabola has equation $y^{2} = 12x$

V_L · Jan 18, 2017

Sorry i attached the wrong answer

But, thank you

Parabola question (1 Viewer)

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