Animated Solution for Mathematics - Conic Sections: The point of intersection of the tangents at the ends of the latus rectum of the parabola $y^2=4x$ is.........

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey into the heart of conic sections. We are going to explore the parabola $y^2=4x$, a curve that defines the path of projectiles and the shape of satellite dishes.

Specifically, we are going to uncover the hidden intersection of the tangents at the ends of its latus rectum. Let us peel back the layers of this problem together.

Every great journey begins with understanding the terrain. We are given the equation $y^2=4x$. To unlock its secrets, we compare it to the standard form of a parabola opening to the right: $y^2=4ax$.

By aligning these two, we see that $4a=4$, which immediately tells us that $a=1$. This parameter $a$ is the DNA of our parabola. It tells us where the focus lies and how 'wide' the curve is.

With $a=1$, we know our focus $S$ is at $(a,0)$, which is $(1,0)$.

Now, let us visualize the latus rectum. It is the focal chord that stands perpendicular to the axis of symmetry. It is the 'widest' part of the parabola near the focus.

The endpoints of this chord are defined by the coordinates $(a,2a)$ and $(a,-2a)$. Substituting our value of $a=1$, we find our two points of interest: $P(1,2)$ and $Q(1,-2)$.

These are the two points where we will construct our tangents. Imagine standing at these points on the curve; we want to know where the lines tangent to the curve at these specific locations will eventually meet.

To find the tangents, we reach for our most reliable tool: the point-form equation of a tangent. For a parabola $y^2=4ax$, the tangent at any point $(x_1,y_1)$ is given by:

This equation is a bridge between the geometry of the curve and the algebra of lines. Let us apply this to our points.

For point $P(1,2)$, we substitute $x_1=1$, $y_1=2$, and $a=1$ into our formula:

Simplifying this, we divide both sides by $2$, yielding the elegant equation:

Now, let us turn our attention to point $Q(1,-2)$. We repeat the process with $x_1=1$, $y_1=-2$, and $a=1$:

Again, dividing by $2$, we get $-y=x+1$, or more simply:

We now have two lines: $y=x+1$ and $y=-x-1$. The question asks for their point of intersection. This is where the algebra becomes satisfying.

We set the two equations equal to each other, or simply add them. If we add the two equations:

With $y=0$, we substitute back into our first equation $y=x+1$ to find $x$:

Our intersection point is $(-1,0)$.

Look closely at the result: $(-1,0)$. Does this coordinate look familiar?

The directrix of the parabola $y^2=4ax$ is defined by the line $x=-a$. Since $a=1$, our directrix is $x=-1$. Our intersection point lies exactly on the directrix!

This is not a coincidence; it is a fundamental geometric truth. The tangents at the extremities of any focal chord of a parabola will always intersect at right angles on the directrix.

You have just derived a powerful theorem through pure algebraic persistence. Remember this, and you will see the beauty of conic sections everywhere you look. Keep practicing, keep questioning, and keep falling in love with the mathematics behind the problems.