If one end of a focal chord of the parabola, y² = 16x is at (1, 4), then the length of this focal ch

Q: Wait, why do we use the formula $L = a(t + \frac{1}{t})^2$ instead of finding the other point $Q$ and using the distance formula?

That is a great question! While you absolutely *can* find the other point $Q$ and use the distance formula, it is much more time-consuming. In competitive exams like JEE, time is your most valuable resource. The formula $L = a(t + \frac{1}{t})^2$ is derived directly from the geometry of the parabola and the properties of focal chords. Memorizing this saves you from tedious algebraic calculations and reduces the chance of making a silly error.

Q: How do I know if $t$ is $1/2$ or $-1/2$?

Excellent observation! The parameter $t$ is determined by the coordinates of the point. We have $2at = y$. Since $a=4$ and the $y$-coordinate of our point is $4$, we have $2(4)t = 4$, which simplifies to $8t = 4$, so $t = 1/2$. If the point had been $(1, -4)$, then $t$ would have been $-1/2$. The sign of $t$ is locked to the sign of the $y$-coordinate.

Q: Can I use the slope form of the chord instead?

Yes, you can! The length of a focal chord with slope $m$ is $4a \csc^2 \theta$, where $m = \tan \theta$. However, this requires you to first find the slope of the line passing through $(1, 4)$ and the focus $(4, 0)$. The parametric approach is generally more direct when you are given a point on the parabola.

Question

If one end of a focal chord of the parabola, $y^2 = 16x$ is at $(1, 4)$, then the length of this focal chord is

Options:

(A) 25

(B) 24

(C) 20

(D) 22

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Focal Chord

Imagine you are standing before the elegant curve of the parabola $y^2 = 16x$. It is a classic, rightward-opening shape, its vertex resting peacefully at the origin $(0, 0)$.

In the world of coordinate geometry, this is not just a curve; it is a locus of points equidistant from a fixed point—the focus—and a fixed line—the directrix. Our mission today is to find the length of a focal chord, a line segment that cuts through this parabola and passes directly through its heart: the focus.

### Decoding the Parabola

First, we must understand the DNA of our parabola. The standard equation for a rightward-opening parabola is $y^2 = 4ax$.

By comparing this to our given equation, $y^2 = 16x$, we see that $4a = 16$. This reveals that $a = 4$.

The focus $S$ of a parabola $y^2 = 4ax$ is always located at $(a, 0)$. Thus, our focus $S$ sits at $(4, 0)$. This point is the anchor for our focal chord.

### The Parametric Key

We are given one end of the chord at the point $P(1, 4)$. To unlock the properties of this point, we use the parametric representation $(at^2, 2at)$.

By equating the $y$-coordinate of our point $P$ to the parametric $y$-coordinate, we get $2at = 4$. Substituting our known value $a = 4$, we have $2(4)t = 4$, which simplifies to $8t = 4$.

Solving for $t$, we find $t = \frac{1}{2}$. This parameter $t$ is the unique identifier for point $P$ on our curve.

### The Focal Chord Magic

Now, we approach the core of the problem. A focal chord is a line segment connecting two points on the parabola that passes through the focus $S$.

Instead of laboriously finding the coordinates of the other end of the chord, we use a beautiful, time-saving formula: the length $L$ of a focal chord with one end at parameter $t$ is given by:

$$L = a\left(t + \frac{1}{t}ight)^2$$

This formula is a gem for any JEE aspirant. It encapsulates the entire geometry of the chord into a simple algebraic expression.

### Final Calculation

With $a = 4$ and $t = \frac{1}{2}$, we are ready to compute. Plugging these values into our formula, we get:

$$L = 4\left(\frac{1}{2} + \frac{1}{1/2}ight)^2$$

The reciprocal of $\frac{1}{2}$ is $2$, so the expression becomes $4(\frac{1}{2} + 2)^2$. Simplifying the sum inside the bracket, we have $\frac{1}{2} + 2 = \frac{5}{2}$.

Now, we square this result:

$$\left(\frac{5}{2}ight)^2 = \frac{25}{4}$$

Finally, we multiply by the $4$ outside:

$$L = 4 	imes \frac{25}{4} = 25$$

The length of the focal chord is **25 units**. It is a clean, satisfying result that highlights the elegance of coordinate geometry.

Animated Solution for Mathematics - Conic Sections: If one end of a focal chord of the parabola, $y^{2} = 16 x$ is at $(1, 4)$ , then the length of this focal chord is

Select Answer:

Visualized Solution

Visualize the Parabola $y^{2} = 16 x$

Identify the Parameter $a$

Locate the Given Point $P (1, 4)$

Find the Parameter $t$

Understand the Focal Chord

Apply the Length Formula

Substitute the Values

Calculate the Sum

Final Computation

Conclusion and Summary

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

Analyzing the Geometry of the Focal Chord

Decoding the Parabola

The Parametric Key

The Focal Chord Magic

Final Calculation

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one end of a focal chord of the parabola, y2=16x is at (1,4), then the length of this focal chord is