The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y

Q: Why do we use parametric coordinates instead of just using $y^2 = 4ax$ directly?

Great question! Using $y^2 = 4ax$ directly forces you to deal with square roots or implicit differentiation, which is messy. Parametric coordinates $Q(at^2, 2at)$ turn the parabola into a single-variable function, effectively 'linearizing' the complexity. It is the standard 'pro' move in coordinate geometry!

Q: How do I know which directrix formula to use for the new parabola?

Always map your derived equation to the standard form $Y^2 = 4AX$. Once you identify the shift in the vertex and the new focal length $A$, the directrix is simply $X = -A$. It is all about pattern matching and recognizing the shifted coordinate system.

Q: Is it a coincidence that the directrix turned out to be the y-axis?

Not at all! It is a beautiful geometric consequence. The midpoint locus essentially 'compresses' the original parabola, shifting its vertex and focal length, which naturally aligns the directrix with the y-axis. It is a sign that your math is perfectly on track!

Question

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^2 = 4ax$ is another parabola with directrix

Options:

(A) $x = -a$

(B) $x = -a/2$

(C) $x = 0$

(D) $x = a/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You see the classic, elegant curve of the parabola $y^2 = 4ax$.

Now, fix your gaze on the focus $S$ at $(a, 0)$. There is a point $Q$ moving along this curve, tracing its path like a dancer on a stage.

We are tasked with finding the path, or the 'locus', of the midpoint $M$ of the segment connecting the fixed focus $S$ to the moving point $Q$.

### The Parametric Bridge

To capture the motion of $Q$, we use the parametric form $Q = (at^2, 2at)$. This reduces the two variables $x$ and $y$ into a single parameter $t$.

Now, let us define our midpoint $M(h, k)$. By the definition of a midpoint, we average the coordinates of $S(a, 0)$ and $Q(at^2, 2at)$:

$$h = \frac{a + at^2}{2}$$

$$k = \frac{0 + 2at}{2}$$

### The Algebraic Surgery

We have our equations, but they are tied to the parameter $t$. To find the locus, we must eliminate $t$.

Looking at the $y$-coordinate:

$$k = \frac{2at}{2} = at \Rightarrow t = \frac{k}{a}$$

Now, we substitute this expression for $t$ into our equation for $h$:

$$h = \frac{a + a(\frac{k}{a})^2}{2}$$

Multiplying by $2$, we get $2h = a + a(\frac{k^2}{a^2})$. The $a$ in the numerator and one $a$ in the denominator cancel out, leaving us with:

$$2h = a + \frac{k^2}{a}$$

### The Reveal of the New Curve

We want to isolate $k^2$ to see the standard form. Multiplying the entire equation by $a$, we get $2ah = a^2 + k^2$.

Rearranging this, we find:

$$k^2 = 2ah - a^2$$

Factoring out $2a$ on the right side, we obtain:

$$k^2 = 2a\left(h - \frac{a}{2}\right)$$

Replacing $(h, k)$ with general coordinates $(x, y)$, we have the equation of our new locus:

$$y^2 = 2a\left(x - \frac{a}{2}\right)$$

### The Final Symmetry

The question asks for the directrix. For a parabola of the form $Y^2 = 4AX$, the directrix is $X = -A$.

Here, our $X$ is $(x - \frac{a}{2})$ and our $4A = 2a$, which means $A = \frac{a}{2}$.

So, the equation for the directrix becomes:

$$x - \frac{a}{2} = -\frac{a}{2}$$

Adding $\frac{a}{2}$ to both sides, we get the final result:

**$x = 0$**

The directrix is the y-axis! It is a stunning, clean result. You have successfully navigated the geometry, the algebra, and the logic.

Animated Solution for Mathematics - Conic Sections: The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^{2} = 4 a x$ is another parabola with directrix

Select Answer:

Visualized Solution

Visualizing the Setup

Defining Coordinates of $S$ and $Q$

Applying the Mid-point Formula

Simplifying the Y-coordinate

Eliminating the Parameter $t$

Simplifying the Locus Equation

Standardizing the Parabola Equation

Identifying the New Parabola's Properties

Finding the Directrix

Final Conclusion

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

Analyzing the Setup

The Parametric Bridge

The Algebraic Surgery

The Reveal of the New Curve

The Final Symmetry

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