Geometry of parabola question (1 Viewer)

gwc · Nov 28, 2016

Hi! Does anyone know how to do this question?

A parabola with vertical axis has its vertex at the origin. Find the equation of the parabola if the line 12x − 4y + 3 = 0 is a tangent.
Answer: y = -3(x^2)
cambridge 9B q.12b

Any help would be much appreciated!

InteGrand · Nov 28, 2016

gwc said:
Hi! Does anyone know how to do this question?

A parabola with vertical axis has its vertex at the origin. Find the equation of the parabola if the line 12x − 4y + 3 = 0 is a tangent.
Answer: y = -3(x^2)
cambridge 9B q.12b

Any help would be much appreciated!

$\noindent The parabola has a vertical axis and vertex at the origin, so its equation is $y = ax^{2}$ for some $a$. We just need to find $a$. Since the line $y = 3x + \frac{3}{4}$ is tangent to the parabola, we must have that the quadratic $ax^{2} - 3x - \frac{3}{4}$ has discriminant $0$ (because the equation formed when finding the common point on the line and parabola is $ax^{2} = 3x + \frac{3}{4} \iff ax^{2} - 3x - \frac{3}{4} = 0$. As we know from the theory in the textbook, the line being a tangent is equivalent to having the discriminant of the LHS quadratic here be $0$.). The discriminant of this quadratic is $9 + 3a$. So we have $9 + 3a = 0$, i.e. $a = -3 $. Hence the parabola has equation $y = -3x^{2}$.$

Geometry of parabola question (1 Viewer)

gwc

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InteGrand

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