Locus of mid point of the portion between the axes of x cos α + y sin α = p where p is constant is (

Q: Wait, why do we replace $h$ and $k$ with $x$ and $y$ at the very end? Isn't that confusing?

That is a fantastic question! In coordinate geometry, $(h, k)$ represents a specific, arbitrary point on the path we are tracking. However, the 'locus' is the set of all such points that satisfy a specific condition. By replacing $(h, k)$ with $(x, y)$, we are essentially saying, 'Any point $(x, y)$ that satisfies this equation is part of the path.' It transforms a specific coordinate into a general rule for the entire curve.

Q: Why can't we just use the distance formula to solve this?

The distance formula is great for finding lengths, but here we are looking for a relationship between the coordinates of a moving point. The distance formula would give you the length of the segment $AB$, which is constant, but it wouldn't tell you how the midpoint moves. We need the midpoint formula to link the moving point to the intercepts.

Q: Is it possible to solve this without using the intercept form?

Absolutely! You could find the intercepts by setting $y=0$ to find $x$, and $x=0$ to find $y$. The intercept form is just a shortcut to get there faster. As long as you correctly identify the coordinates of $A$ and $B$, the method remains valid. Always choose the path that feels most intuitive to you!

Question

Locus of mid point of the portion between the axes of $x \cos \alpha + y \sin \alpha = p$ where $p$ is constant is

Options:

(A) $x^2 + y^2 = \frac{4}{p^2}$

(B) $x^2 + y^2 = 4p^2$

(C) $\frac{1}{x^2} + \frac{1}{y^2} = \frac{2}{p^2}$

(D) $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation $x \cos \alpha + y \sin \alpha = p$ represents the **Normal Form** of a straight line. In this expression, $p$ is the constant perpendicular distance from the origin to the line, and $\alpha$ is the angle the perpendicular makes with the positive $x$-axis.

As $\alpha$ varies, the line rotates and shifts while maintaining a constant distance $p$ from the origin. Our objective is to determine the locus of the midpoint of the segment intercepted by this line on the coordinate axes.

### Phase 1: The Intercepts

To find the endpoints of the segment, we identify the points where the line intersects the $x$-axis (point $A$) and the $y$-axis (point $B$). We transform the normal form into the intercept form, $\frac{x}{a} + \frac{y}{b} = 1$, by dividing the original equation by $p$:

$$\frac{x \cos \alpha}{p} + \frac{y \sin \alpha}{p} = 1$$

Rearranging this into the standard intercept form yields:

$$\frac{x}{\left( \frac{p}{\cos \alpha} \right)} + \frac{y}{\left( \frac{p}{\sin \alpha} \right)} = 1$$

Consequently, the coordinates of the endpoints are $A\left( \frac{p}{\cos \alpha}, 0 \right)$ and $B\left( 0, \frac{p}{\sin \alpha} \right)$.

### Phase 2: The Midpoint's Dance

Let the midpoint of segment $AB$ be $M(h, k)$. Applying the midpoint formula, we calculate the coordinates as the average of the endpoints:

$$h = \frac{\frac{p}{\cos \alpha} + 0}{2} = \frac{p}{2 \cos \alpha}$$

$$k = \frac{0 + \frac{p}{\sin \alpha}}{2} = \frac{p}{2 \sin \alpha}$$

We now have a system of equations defining the position of $M$ in terms of the parameter $\alpha$. To find the locus, we must eliminate $\alpha$ to establish a direct relationship between $h$ and $k$.

### Phase 3: The Elimination Strategy

We isolate the trigonometric functions from our midpoint equations:

$$\cos \alpha = \frac{p}{2h}, \quad \sin \alpha = \frac{p}{2k}$$

We utilize the fundamental trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$ to bridge these expressions. Substituting our values into the identity gives:

$$\left( \frac{p}{2k} \right)^2 + \left( \frac{p}{2h} \right)^2 = 1$$

Expanding this expression, we obtain:

$$\frac{p^2}{4k^2} + \frac{p^2}{4h^2} = 1$$

### Phase 4: The Final Reveal

To simplify, we divide the entire equation by $\frac{p^2}{4}$:

$$\frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2}$$

Replacing the specific point $(h, k)$ with the general variables $(x, y)$, we arrive at the final locus of the midpoint:

$$\mathbf{\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}}$$

This symmetric equation elegantly describes the path traced by the midpoint as the line slides. By navigating the geometry and systematically eliminating the parameter, we have reduced a dynamic system to a static, fundamental curve.

Animated Solution for Mathematics - Straight Lines: Locus of mid point of the portion between the axes of $x cos α + y sin α = p$ where $p$ is constant is

Select Answer:

Visualized Solution

Visualizing the Line $x cos α + y sin α = p$

Converting to Intercept Form

Coordinates of Intercepts $A$ and $B$

Introducing the Midpoint $M$

Applying Midpoint Formula for $x$ -coordinate

Applying Midpoint Formula for $y$ -coordinate

Isolating the Parameter $α$

The Trigonometric Identity

Substituting into the Identity

Expanding the Squares

Rearranging to Final Form

Final Locus Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Locus of mid point of the portion between the axes of xcosα+ysinα=p where p is constant is