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Problem statement:

Two tangents to the parabola $y=x^2$ are perpendicular. Prove that the intersection between the tangents lies on the line $y=-\frac{1}{4}$.

Solution:

Two tangents $y_1=k_1x+m$, $y_2=k_2x+n$ are perpendicular if $k_1k_2=-1$ so we can reduce the second tangent to $y_2=-\frac{1}{k_2}x+n$. Before finding the intersection which is $y_1=y_2$ I believe we need another condition that relates $y_1,y_2$ to $y=x^2$. I'm sure it has to do with the derivative but I am stuck.

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The tangents to the points $(x_1,x_1^2)$, $(x_2,x_2^2)$, with $x_1,x_2\neq 0$, $x_1\neq x_2$ are

$$y =2x_i x -x^2_i, $$

for $i=1,2$. In particular, the slope is found using $y'(x)=2x$. As the 2 tangents are supposed to be perpendicular, then

$$4x_1x_2=-1, $$

or $x_1x_2=-\frac{1}{4}$. All you need is to find the point of intersection between the lines: subtract $y =2x_2 x -x^2_2 $ from $y =2x_1 x -x^2_1 $ arriving at

$$x=\frac{1}{2}(x_2+x_1). $$

Plug such coordinate into $y =2x_2 x -x^2_2 $ arriving at

$$y=x_2(x_2+x_1)-x^2_2$$

or

$$y=x_2x_1=-\frac{1}{4}$$

as claimed.

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Hints and facts here:

  • $y-y_0=2x_0(x-x_0)$ is taken as a line tangent to $f(x)$ at $(x_0,y_0)$.

  • $y-y_1=2x_1(x-x_1)$ is taken as a line tangent to $f(x)$ at $(x_1,y_1)$.

  • $y_1=x_1^2,~~y_0=x_0^2$.

  • $2x_1=(-1/2x_0)$.

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  • $\begingroup$ You've been great with hints! $+_+^++_+^++$ $\endgroup$ – Namaste Aug 22 '13 at 12:32
  • $\begingroup$ @amWhy: Good time Amy. :-) $\endgroup$ – mrs Aug 22 '13 at 12:32
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Step 1: The tangent to the parabola at the point $x_0$ has the form

$$f(x)=2xx_0-x_0^2$$

(you know the slope by differentiation and one point on the tangent, namely $(x_0,x_0^2)$).

Step 2: The tangents at $x_0$ and $x_1$ will intersect at $(\frac{x_0+x_1}{2},x_0x_1)$

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A parabola can be defined as the set of all points that have the same distance from a point (the focus) and a line (the directrix).

Note that $y = -\frac 1 4$ is the directrix of the parabola $y = x^2$. So we are to prove that the two tangents intersect on the directrix.

A short Lemma: Let $g$ be the directrix line and $F$ the focus of the parabola. Let $l$ be a tangent to the parabola and let $P$ be the point where it intersects the parabola. Let $A$ the orthogonal projection of $P$ onto $g$. Then it is easy to see from above definition of a parabola that $l$ is the perpendicular bisector of the line segment $AF$ (simply because this perpendicular bisector intersects the parabola at the point $P$ and in no other point). (End of Lemma)

In our case we have two tangents $l_1$ and $l_2$ which are perpendicular. Let $A_1$ and $A_2$ be the corresponding points on $g$ (as in the lemma). Then $A_1 F$ and $A_2 F$ must also be perpendicular. So the angle $A_1 F A_2$ is a right angle. Therefore the circumcenter of the triangle $A_1 F A_2$ (which is where the lines $l_1$ and $l_2$ intersect) is on the line segment $A_1 A_2$ (which is part of the line $g$.)

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