Through the vertex O of parabola y² = 4x, chords OP and OQ are drawn at right angles to one another.

Q: Wait, why do we use $t_1 t_2 = -4$ instead of just solving for $t_1$ and $t_2$?

That is a fantastic question! The beauty of this problem lies in the fact that we do not need to know the specific values of $t_1$ and $t_2$. Because the chords can rotate freely while maintaining their perpendicularity, $t_1$ and $t_2$ are variables. However, their product is invariant—it is a constant. By using this invariant, we eliminate the dependency on the specific positions of $P$ and $Q$, allowing us to prove a property that holds true for *all* such chords.

Q: Why does the locus of the midpoint turn out to be another parabola?

It is not a coincidence! The midpoint of a chord is a geometric average. When you average the coordinates of two points on a parabola, the resulting path often retains the quadratic nature of the original curve. In this case, the transformation $y^2 = 2(x - 4)$ shows that the locus is simply a scaled-down version of the original parabola, shifted to start at the fixed point $(4, 0)$. It is a beautiful example of geometric self-similarity.

Q: I got stuck on the algebra for the locus. How do I eliminate $t_1$ and $t_2$?

The trick is to treat the midpoint coordinates $(h, k)$ as the target. We have $k = t_1 + t_2$ and $h = \frac{t_1^2 + t_2^2}{2}$. We need to relate these to the product $t_1 t_2 = -4$. The algebraic identity $(t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2$ is your bridge. By substituting $k^2$ for the left side and $2h$ for the sum of squares, you naturally eliminate the parameters and are left with a clean equation in $h$ and $k$.

Question

Through the vertex $O$ of parabola $y^2 = 4x$, chords $OP$ and $OQ$ are drawn at right angles to one another. Show that for all positions of $P$, $PQ$ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of $PQ$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not merely solving a problem; we are witnessing a beautiful, hidden symmetry in the geometry of the parabola.

Imagine standing at the origin of the Cartesian plane, the vertex of the parabola $y^2 = 4x$. You are holding a flashlight, and you are drawing two beams of light, $OP$ and $OQ$, that strike the parabola at points $P$ and $Q$.

The constraint is simple yet profound: these beams must be perpendicular. As you rotate your flashlight, the points $P$ and $Q$ dance along the curve, but the line connecting them, the chord $PQ$, behaves in a way that is almost magical. It always pivots through a fixed point on the axis.

### Phase 1: The Parametric Power-Up

In coordinate geometry, the choice of coordinates is everything. If we stick to standard Cartesian coordinates $(x, y)$, we will quickly find ourselves drowning in square roots and messy algebraic expressions.

Instead, let us embrace the elegance of the parametric form. For the parabola $y^2 = 4ax$, where $a=1$, any point can be described by a single parameter $t$ as $(t^2, 2t)$.

Let point $P$ be defined by the parameter $t_1$, giving us coordinates $(t_1^2, 2t_1)$, and let point $Q$ be defined by $t_2$, giving us $(t_2^2, 2t_2)$. By doing this, we have reduced a two-dimensional problem into a one-dimensional parameter space.

### Phase 2: The Orthogonality Constraint

Now, let us apply the core constraint: $OP \perp OQ$. In the language of slopes, this means the product of the slopes of $OP$ and $OQ$ must be $-1$.

The slope of $OP$ is:
$$\frac{2t_1 - 0}{t_1^2 - 0} = \frac{2}{t_1}$$

Similarly, the slope of $OQ$ is $\frac{2}{t_2}$. When we multiply these, we get:
$$\left( \frac{2}{t_1} \right) \cdot \left( \frac{2}{t_2} \right) = -1$$

This simplifies to $\frac{4}{t_1 t_2} = -1$, or more elegantly, $t_1 t_2 = -4$. This constant relationship is the heartbeat of our problem and the key that will unlock the rest of the mystery.

### Phase 3: The Chord Equation

With our parameters locked, let us find the equation of the line $PQ$. The slope of the chord $PQ$ is:
$$\frac{2t_2 - 2t_1}{t_2^2 - t_1^2} = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$$

Using the point-slope form with point $P$, we write:
$$y - 2t_1 = \frac{2}{t_1 + t_2}(x - t_1^2)$$

Multiplying through by $(t_1 + t_2)$, we get $y(t_1 + t_2) - 2t_1(t_1 + t_2) = 2(x - t_1^2)$. Expanding this, we find $y(t_1 + t_2) - 2t_1^2 - 2t_1 t_2 = 2x - 2t_1^2$.

Notice the beautiful cancellation of $-2t_1^2$ on both sides! We are left with the master equation:
$$y(t_1 + t_2) - 2x = 2t_1 t_2$$

### Phase 4: The Fixed Point Revelation

We want to see where this chord cuts the axis of the parabola. The axis is the $x$-axis, where $y = 0$.

Substituting $y = 0$ into our equation, we get $-2x = 2t_1 t_2$, which simplifies to $-x = t_1 t_2$. Since we know $t_1 t_2 = -4$, we substitute this in to get $-x = -4$, or $x = 4$.

The chord $PQ$ always passes through the fixed point $(4, 0)$, regardless of the values of $t_1$ and $t_2$. We have successfully proven the first part of our quest.

### Phase 5: The Locus of the Midpoint

Finally, let us find the locus of the midpoint $R(h, k)$ of the chord $PQ$. The coordinates are:
$$h = \frac{t_1^2 + t_2^2}{2}, \quad k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2$$

We need to eliminate $t_1$ and $t_2$. We use the identity $(t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2$. Substituting our expressions, we get:
$$k^2 = 2h + 2(-4)$$

This simplifies to $k^2 = 2h - 8$, or $k^2 = 2(h - 4)$. Replacing $(h, k)$ with $(x, y)$, we arrive at the final locus:
$$\mathbf{y^2 = 2(x - 4)}$$

This is a parabola, shifted to the right, with its vertex at the fixed point $(4, 0)$. We have successfully mapped the path of the midpoint, revealing a new, harmonious geometric structure born from the original parabola.

Visualized Solution

Visualizing the Parabola and Chords

Parametric Coordinates of $P$ and $Q$

Applying the Perpendicularity Condition

Simplifying the Orthogonality Relation

Deriving the Equation of Chord $P Q$

Simplifying the Chord Equation

Finding the Intersection with the Axis

Setting up the Midpoint Coordinates

Eliminating $t_{1}$ and $t_{2}$

The Locus of the Midpoint

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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