Animated Solution for Mathematics - Conic Sections: Let a tangent to the curve $y^2=24x$ meet the curve $xy=2$ at the points A and B. Then the mid points of such line segments AB lie on a parabola with the

Imagine you are standing on a coordinate plane, watching two distinct mathematical entities: the parabola $y^2=24x$ and the rectangular hyperbola $xy=2$. We are looking for a line that is a tangent to the parabola and, at the same time, acts as a chord for the hyperbola.

Our goal is to find the path—the locus—traced by the midpoint of this chord as the line moves.

The Parabola's Tangent

Let us start with the parabola $y^2=24x$. Comparing this to the standard form $y^2=4ax$, we immediately see that $4a=24$, which gives us $a=6$.

Any tangent to this parabola can be described using a parameter $t$. The equation of this tangent is $ty=x+a{t}^2$.

Substituting our value of $a$, we get:

This is our first anchor. It is a line that depends on $t$. As $t$ changes, the line slides along the parabola, always staying tangent.

The Hyperbola's Chord

Now, shift your focus to the hyperbola $xy=2$. We are interested in a chord of this hyperbola whose midpoint is $(h,k)$.

We use the 'magic' of coordinate geometry: the $T=S_1$ formula. For any conic $S=0$, the chord with midpoint $(h,k)$ is given by $T=S_1$.

For our hyperbola $xy-2=0$, $T$ becomes $\frac{xk+yh}{2}-2$ and $S_1$ becomes $hk-2$. Setting $T=S_1$, we get:

The constant $-2$ cancels out beautifully, leaving us with:

This is the equation of the chord in terms of the midpoint $(h,k)$.

The Bridge

We have two equations for the same line: $x-ty+6t^2=0$ and $kx+hy-2hk=0$. Since they represent the same line, the ratios of their coefficients must be equal.

We write:

This proportionality is the bridge between the parabola's parameter $t$ and the hyperbola's midpoint $(h,k)$.

The Locus

Now, we perform the algebraic surgery. From the first two terms, $\frac{1}{k}=\frac{-t}{h}$, we find:

From the first and third terms, $\frac{1}{k}=\frac{6t^2}{-2hk}$, we simplify to get $-2h=6t^2$, or:

Substituting $t=-\frac{h}{k}$ into $3t^2=-h$, we get $3(-\frac{h}{k}{)}^2=-h$, which simplifies to:

Canceling $h$ (assuming $h eq 0$), we get $3h=-k^2$, or $k^2=-3h$. Replacing $(h,k)$ with $(x,y)$, we arrive at the final locus: $y^2=-3x$

This is a parabola opening to the left. We have successfully navigated the geometry and found our answer. The beauty of this problem lies in how two seemingly unrelated curves dictate the path of a single point.