Locus of the mid-points of the portion of the line $x\sin\theta + y\cos\theta = p$ intercepted between the axes is ____.

> [!TLDR] > The key idea is to find the intercepts of the given line on the axes, then express the coordinates of the midpoint in terms of $\theta$, and eliminate $\theta$ to get the locus. The final locus is $4x^2 y^2 = p^2 (x^2 + y^2)$. We start with the line equation $x\sin\theta + y\cos\theta = p$. This is a line that depends on the parameter $\theta$. For each $\theta$, the line cuts the x-axis and y-axis at two points. The segment between these intercepts has a midpoint. As $\theta$ varies, this midpoint traces a curve — that curve is the locus we need. The natural approach: find the intercepts, write the midpoint coordinates in terms of $\theta$, then eliminate $\theta$ to get a relation between $x$ and $y$ that holds for all midpoints. --- 1. **Find the x-intercept** Set $y = 0$ in the line equation: $$x\sin\theta = p \quad\Rightarrow\quad x = \frac{p}{\sin\theta}$$ So the x-intercept is $A\left(\frac{p}{\sin\theta},\, 0\right)$. 2. **Find the y-intercept** Set $x = 0$: $$y\cos\theta = p \quad\Rightarrow\quad y = \frac{p}{\cos\theta}$$ So the y-intercept is $B\left(0,\, \frac{p}{\cos\theta}\right)$. 3. **Midpoint of segment AB** Let the midpoint be $M(h, k)$. Then: $$h = \frac{\frac{p}{\sin\theta} + 0}{2} = \frac{p}{2\sin\theta}$$ $$k = \frac{0 + \frac{p}{\cos\theta}}{2} = \frac{p}{2\cos\theta}$$ So we have: $$h = \frac{p}{2\sin\theta}, \quad k = \frac{p}{2\cos\theta}$$ 4. **Eliminate $\theta$** From the above: $$\sin\theta = \frac{p}{2h}, \quad \cos\theta = \frac{p}{2k}$$ Use the identity $\sin^2\theta + \cos^2\theta = 1$: $$\left(\frac{p}{2h}\right)^2 + \left(\frac{p}{2k}\right)^2 = 1$$ $$\frac{p^2}{4h^2} + \frac{p^2}{4k^2} = 1$$ Multiply through by $4h^2 k^2$: $$p^2 k^2 + p^2 h^2 = 4h^2 k^2$$ $$p^2 (h^2 + k^2) = 4h^2 k^2$$ …

Locus of the mid-points of the portion of the line xsinθ + ycosθ = p intercepted between the axes is __.

NCERT Exemplar · Q47

x\sin\theta + y\cos\theta = p

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Concept understanding — Coordinate Axes Equations

Coordinate Axes Equations

Imagine a blank sheet of paper. You want to describe exactly where a point is. The most natural way is to draw two perpendicular lines — one horizontal, one vertical — and measure distances from them. Those two lines are the coordinate axes.

The horizontal line is the x-axis. The vertical line is the y-axis. Where they cross is the origin, the point $(0,0)$ .

Now, here's the key idea: every point on the x-axis has a y-coordinate of 0. Why? Because to be on that horizontal line, you haven't moved up or down at all — your vertical distance from the axis is zero. Similarly, every point on the y-axis has an x-coordinate of 0 — you haven't moved left or right from the vertical line.

That's the entire intuition. Let's make it precise.

The Equations

\text{x-axis: } y = 0

That's it. These are the equations of the coordinate axes.

$y = 0$ means: "all points where the y-coordinate is zero, regardless of x." That's the entire horizontal line through the origin.
$x = 0$ means: "all points where the x-coordinate is zero, regardless of y." That's the entire vertical line through the origin.

Watch out

A common mistake is to think $y=0$ means "the point $(0,0)$ ". No — it means every point with $y=0$ , like $(5,0)$ , $(-3,0)$ , $(100,0)$ , etc. It's a whole line, not a single point.

Why This Matters

These two equations are the foundation for everything in coordinate geometry. Every other line, curve, or shape is described relative to these axes. For example:

The line $y = 2$ is a horizontal line parallel to the x-axis, shifted up by 2 units.
The line $x = -3$ is a vertical line parallel to the y-axis, shifted left by 3 units.

So when you see $y = 0$ or $x = 0$ , think: "the original, unshifted axes themselves." …

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The key idea is to find the intercepts of the given line on the axes, then express the coordinates of the midpoint in terms of $\theta$ , and eliminate $\theta$ to get the locus. The final locus is $4x^2 y^2 = p^2 (x^2 + y^2)$ .

We start with the line equation $x\sin\theta + y\cos\theta = p$ . This is a line that depends on the parameter $\theta$ . For each $\theta$ , the line cuts the x-axis and y-axis at two points. The segment between these intercepts has a midpoint. As $\theta$ varies, this midpoint traces a curve — that curve is the locus we need.

The natural approach: find the intercepts, write the midpoint coordinates in terms of $\theta$ , then eliminate $\theta$ to get a relation between $x$ and $y$ that holds for all midpoints.

Find the x-intercept Set $y = 0$ in the line equation:

x\sin\theta = p \quad\Rightarrow\quad x = \frac{p}{\sin\theta}

So the x-intercept is $A\left(\frac{p}{\sin\theta},\, 0\right)$ .

Find the y-intercept Set $x = 0$ :

y\cos\theta = p \quad\Rightarrow\quad y = \frac{p}{\cos\theta}

So the y-intercept is $B\left(0,\, \frac{p}{\cos\theta}\right)$ .

Midpoint of segment AB Let the midpoint be $M(h, k)$ . Then:

h = \frac{\frac{p}{\sin\theta} + 0}{2} = \frac{p}{2\sin\theta}

k = \frac{0 + \frac{p}{\cos\theta}}{2} = \frac{p}{2\cos\theta}

So we have:

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

Eliminate $\theta$ From the above:

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

Use the identity $\sin^2\theta + \cos^2\theta = 1$ :

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

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

Multiply through by $4h^2 k^2$ :

p^2 k^2 + p^2 h^2 = 4h^2 k^2

$p^2 (h^2 + k^2) = 4h^2 k^2$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$(x, y, z)$
›Reveal solutionSolution
The sum of the squares of perpendicular distances from a point to the coordinate axes equals twice the square of the distance from the origin, so $k = 2$ .

We are asked: For a point $(x, y, z)$ , the sum of the squares of its perpendicular distances to the three coordinate axes equals $k$ times the square of its distance from the origin. Find $k$ .

Concept & Intuition

The "perpendicular distance" from a point to an axis is the shortest distance from the point to any point on that axis. For the $x$ -axis, the foot of the perpendicular is $(x, 0, 0)$ , so the distance is $\sqrt{y^2 + z^2}$ . Similarly for the $y$ -axis and $z$ -axis. The distance from the origin is $\sqrt{x^2 + y^2 + z^2}$ . The problem asks how these quantities relate — specifically, the sum of the squares of those three axis-distances compared to the square of the origin-distance.

Step-by-step reasoning

Write the perpendicular distances to each axis

Distance to $x$ -axis: $\sqrt{y^2 + z^2}$

Distance to $y$ -axis: $\sqrt{x^2 + z^2}$

Distance to $z$ -axis: $\sqrt{x^2 + y^2}$

Square each distance

Square of distance to $x$ -axis: $y^2 + z^2$

Square of distance to $y$ -axis: $x^2 + z^2$

Square of distance to $z$ -axis: $x^2 + y^2$

Sum these squares

$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 2x^2 + 2y^2 + 2z^2$

Compare with the square of the distance from the origin …
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Q.Locus of the mid-points of the portion of the line xsin⁡θ+ycos⁡θ=px\sin\theta + y\cos\theta = pxsinθ+ycosθ=p intercepted between the axes is ____.

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Q.Locus of the mid-points of the portion of the line $x\sin\theta + y\cos\theta = p$ intercepted between the axes is ____.

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