# Application of derivative - tangents to latus rectum

Drawn thru the focus of parabola is a chord perpendicular to the axis of the parabola. Two tangent lines are drawn through the points of intersection of the chord and the parabola. Prove that the tangent intersect at right angles.

We know the chord perpendicular the axis of the parabola is latus rectum of parabola.

Let us consider standard equation of parabola $y^2 =4ax$, Equation of tangent at point P($x_1,y_1)$ to the parabola $y^2=4ax$ is $yy_1 =2a(x+x_1)$

I am not able to get the desired result by using these.. please guide further... thanks..

• You didn't use the fact that $x_1$ is the x-coordinate of the focus. Besides, you need two tangents. What's the equation of the other tangent. May 8 '13 at 9:32
• $yy_2 =2a(x+x_2)$ is the equation of other tangent. May 8 '13 at 12:14

A parabola whose vertex is at $(0,0)$ and opens upward is described by $\{(x,y)\in \mathbb R^2: 4py=x^2\}$ where the focus is at the point $(0,p)$ and the axis of the parabola is the $y$ axis. Therefore, a chord perpendicular to the axis of the parabola and passing through the focus is simply the line described by $\{(x,y)\in \mathbb R^2: y=p\}$.
This line intersects the parabola at $4p^2 = x^2$ or $x=\pm 2p$. The derivative of $y=\frac{x^2}{4p}$ is $y^{\prime} = \frac{x}{2p}$ which is $1$ at $x=2p$ and $-1$ at $x=-2p$. Since these slopes are the negative reciprocals of each other they are perpendicular.