Question

The locus of the midpoint of the intercept of the variable line \(\mathrm{x}\) cosa \(+\) ysina \(=\mathrm{p}\) between the coordinate axes is (a) \(x^{-2}+y^{-2}=p^{-2}\) (b) \(x^{-2}+y^{-2}=2 p^{-2}\) (c) \(x^{-2}+y^{-2}=4 p^{-2}\) (d) none of these

Short answer

The locus of the midpoint of the intercept of the variable line \(x\cos{a} + y\sin{a} = p\) between the coordinate axes is given by the equation \(x^{-2}+y^{-2} = 2 p^{-2}\), which corresponds to option (b).

Step-by-step solution

Find the x-intercept

In order to find the x-intercept, we need to set y = 0 and then solve for x in the given line equation. The given equation is \(x\cos{a} + y\sin{a} = p\). When we put y = 0, we get: \(x\cos{a} = p\) \(x = \dfrac{p}{\cos{a}}\)

Find the y-intercept

To determine the y-intercept, set x = 0 and solve for y in the line equation: \(y\sin{a} = p\) \(y = \dfrac{p}{\sin{a}}\)

Find the midpoint of the intercept

Now that we have the coordinates of the x-intercept and y-intercept, we can calculate the midpoint M by averaging the x-coordinates and the y-coordinates: \(M(\frac{\frac{p}{\cos{a}}+0}{2}, \frac{\frac{p}{\sin{a}}+0}{2})\) This simplifies to: \(M(\frac{p}{2\cos{a}},\frac{p}{2\sin{a}})\)

Find the locus of the midpoint

Let the midpoint be represented by the coordinates (x,y). Then, \(x = \frac{p}{2\cos{a}}\) and \(y = \frac{p}{2\sin{a}}\) Now, we eliminate p from the above equations to obtain the locus: \(\dfrac{2x\cos{a}}{2y\sin{a}} = \frac{x}{y}\) Squaring both sides: \(x^2 = y^2 \cos ^2 {a}\) and \(y^2 = x^2\sin^2{a}\) Adding these two equations: \(x^2(1+\sin^2{a}) = y^2(1+\cos^2{a})\) As we know that \(\sin^2{a} + \cos^2{a} = 1\), we have: \(x^2 = y^2\) \(\Rightarrow x^2+y^2 = 2x^2\) \(\Rightarrow \frac{1}{x^2}+\frac{1}{y^2} = \frac{2}{p^2}\) So, the locus of the midpoint is given by the equation: \(x^{-2}+y^{-2} = 2 p^{-2}\) Therefore, the answer is (b) \(x^{-2}+y^{-2}=2 p^{-2}\).

Key concepts

Understanding Coordinate Geometry

Coordinate geometry is a branch of mathematics that allows us to analyze geometrical shapes using a set of coordinates. It combines algebra and geometry to enable the representation of lines, curves, and figures on a plane. The basic elements involve the x-axis (horizontal) and y-axis (vertical), which intersect at a point called the origin.

Every point on the plane can be described by an ordered pair of numbers (x, y), known as coordinates. The first number, x, defines the position along the horizontal axis, while the second number, y, indicates the position on the vertical axis. The equation of a line in the coordinate plane can provide valuable information, such as the line's slope, intercepts, and midpoint, which are important for solving locus problems in JEE Maths examinations.

Deciphering the Locus of Midpoint

The locus of a point is the collection of all positions that satisfy a given condition. When dealing with the locus of a midpoint in coordinate geometry, we're looking for a set of all possible midpoints under specific constraints.

In JEE Maths problems, the midpoint of a line segment is obtained by averaging the x-coordinates and the y-coordinates of the endpoints. When these endpoints move following a certain rule, the midpoint traces a path – this path is what we call the locus. For example, the locus of the midpoint of a line segment whose ends lie on the coordinate axes can be expressed by a geometrically significant equation, capturing the essence of the path that the midpoint traces as the line moves.

Analyzing Line Intercepts

Line intercepts are points where a line crosses the coordinate axes. Specifically, the x-intercept is the point at which the line crosses the x-axis, and the y-intercept is where it crosses the y-axis. These points are useful in characterizing the line, particularly when discussing the locus of a midpoint.

To find an intercept, one must set the opposite variable to zero and solve for the remaining variable. For instance, setting y = 0 yields the x-intercept, and setting x = 0 gives us the y-intercept. When working with variable lines in JEE Maths, understanding how to find and use these intercepts is crucial, as they often serve as endpoints from which midpoints, and in turn their loci, are calculated.