locus on the plane

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#1 h Oct 9, 2007, 10:09 PM • 2 Y

Y by Adventure10, Mango247

Find the locus of the points $M$ in the plane $Oxy$ such that the tangents from $M$ to $y=x^2$ are perpendicular.

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#2 h Oct 10, 2007, 12:20 PM • 2 Y

Y by Adventure10, Mango247

ElChapin wrote:

Find the locus of the points $M$ in the plane $Oxy$ such that the tangents from $M$ to $y = x^2$ are perpendicular.

Let $y=mx+c$ is a tangent to the curve $y=x^2$ . So, the equation $mx+c=x^2$ must have equal roots. This gives us $m^2+4c=0$ . Hence we have $c=-\frac {m^2}{4}$ . So any tangent to the given curve can be written as $y=mx-\frac {m^2}{4}$
Let this tangent passes through $M(h, k)$ a point in the plane.
Then we have $k=mh-\frac {m^2}{4}$
This is a quadratic equation in m and will have two roots for m. Let the roots are $m_1, m_2$ . If the tangents through M are perpendicular, the product of the slopes must be equal to $-1$ . So we should have $k=-\frac {1}{4}$ . Hence the locus of M is the line $y=-\frac {1}{4}$ which is the directrix of the parabola given.

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