Question

The locus of mid points of the segment intercepted between the axes by the line xseca \(+\) ytana \(=p\) is \(\ldots \ldots\) (a) \(\left[\mathrm{p}^{2} /\left(4 \mathrm{x}^{2}\right)\right]=1+\left[\mathrm{p}^{2} /\left(4 \mathrm{y}^{2}\right)\right]\) (b) \(\left(\mathrm{x}^{2} / \mathrm{p}^{2}\right)+\left(\mathrm{y}^{2} / \mathrm{p}^{2}\right)=4\) (c) \(\left(\mathrm{p}^{2} / \mathrm{x}^{2}\right)=1+\left(\mathrm{p}^{2} / \mathrm{y}^{2}\right)\) (d) \(\left[\mathrm{p}^{2} /\left(4 \mathrm{x}^{2}\right)\right]+\left[\mathrm{p}^{2} /\left(4 \mathrm{y}^{2}\right)\right]=1\)

Short answer

The locus of midpoints of the segment intercepted between the axes by the line xseca \(+\) ytana \(=p\) is: \(\frac{p^2}{4x^2} + \frac{p^2}{4y^2} = 1\)

Step-by-step solution

Determine the points where the line intersects the axes

To find the points where the line intersects the axes, we can set x to 0 and find y, or set y to 0 and find x. The line's equation is xsecα \(+\) ytanα \(=p\). When x=0, the equation becomes: \( ytan\alpha = p \). So, \( y = \frac{p}{tan\alpha} \). When y=0, the equation becomes: \( xsec\alpha = p \). So, \( x = \frac{p}{sec\alpha} = pcos\alpha \). The points where the line intersects the axes are: \((0, \frac{p}{tan\alpha})\) and \((pcos\alpha, 0)\).

Find the midpoint of the line segment

To find the midpoint of the line segment between the points of intersection, we need to average their x and y coordinates respectively. Let the midpoint be (h, k). The x-coordinate of the midpoint, h, is given by: \( h = \frac{0 + pcos\alpha}{2} = \frac{p}{2}cos\alpha\). The y-coordinate of the midpoint, k, is given by: \( k = \frac{\frac{p}{tan\alpha} + 0}{2} = \frac{p}{2tan\alpha}\).

Rewrite the midpoint coordinates in terms of x and y

Since we need to find the locus, we will rewrite h and k in terms of x and y. From the given midpoint coordinates: \(x = \frac{p}{2}cos\alpha \Rightarrow cos\alpha = \frac{2x}{p}\). \(y = \frac{p}{2tan\alpha} \Rightarrow tan\alpha = \frac{p}{2y}\).

Express tanα in terms of cosα

To express tanα in terms of cosα, we will use the fact that \(tan\alpha = \frac{sin\alpha}{cos\alpha}\). Recall that \(sin^2\alpha + cos^2\alpha = 1\) and so, \(sin\alpha = \sqrt{1 - cos^2\alpha}\). Now, \(tan\alpha = \frac{sin\alpha}{cos\alpha} = \frac{\sqrt{1 - cos^2\alpha}}{cos\alpha}\). Substitute the expression for cosα found in Step 3: \(tan\alpha = \frac{\sqrt{1 - (\frac{2x}{p})^2}}{\frac{2x}{p}}\). Now, replace tanα in the expression found in Step 3: \(y = \frac{p}{2 \cdot \frac{\sqrt{1 - (\frac{2x}{p})^2}}{\frac{2x}{p}}}\).

Simplify the expression and identify the locus

Simplify the expression for y: \(y = \frac{p}{2} \cdot \frac{\frac{2x}{p}}{\sqrt{1 - (\frac{2x}{p})^2}}\) \(y = \frac{px}{\sqrt{p^2 - 4x^2}}\) Squaring both sides to get rid of the square root: \(y^2 = \frac{p^2x^2}{p^2 - 4x^2}\) Now, multiply both sides by \((p^2 - 4x^2)\): \(y^2 \cdot (p^2 - 4x^2) = p^2x^2\) Divide by \(p^2\): \(\frac{y^2}{p^2} \cdot (p^2 - 4x^2) = x^2\) Now, divide both sides by \(4x^2y^2\) and rearrange: \(\frac{p^2}{4x^2} + \frac{p^2}{4y^2} = 1\) Looking at the given options, we can see that option (d) matches the final expression we found: (d) $\left[\mathrm{p}^{2} /\left(4 \mathrm{x}^{2}\right)\right]+\left[\mathrm{p}^{2} /\left(4 \mathrm{y}^{2}\right)\right]=1$

Key concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that involves the study of geometric figures using a coordinate system. This approach allows us to describe geometric shapes algebraically and to solve geometric problems with equations and algebraic techniques. The coordinate system most commonly used is the Cartesian coordinate system, which represents points in space with numerical coordinates based on two perpendicular axes typically labeled x (horizontal) and y (vertical).

Key concepts within coordinate geometry include the distance between points, the midpoint of a line segment, and the slope of a line. When dealing with the locus of midpoints, the goal is to find a set of all points that satisfy certain conditions. In our example, we are interested in the locus of midpoints for a line segment intercepted between the axes by a line with a given trigonometric equation. The ability to translate a geometric problem into algebraic expressions and vice versa is crucial in coordinate geometry.

Intersection of Line with Axes

The intersection of a line with the axes in coordinate geometry refers to the points where the line crosses the x-axis and the y-axis. These intersections are called x-intercept and y-intercept respectively. To find these points, one must set one of the variables (x or y) to zero and solve for the other.

For example, in our exercise we use the trigonometric form of the line's equation to find the points where the line intersects the axes. Setting x to 0 gives us the y-intercept, and setting y to 0 gives us the x-intercept. The midpoints of line segments between the axes are of particular interest because they can give us insight into the properties of the line and can be used in various applications, such as determining the locus of a set of points.

Equation of a Line in Trigonometric Form

An equation of a line in trigonometric form involves trigonometric functions of an angle often associated with the slope or orientation of the line. Specifically, in our example, the line is expressed using secant (sec) and tangent (tan) of an angle α, which are trigonometric functions derived from the cosine and sine functions respectively.

For lines in trigonometric form such as xsecα + ytana = p, one can use trigonometric identities to manipulate and simplify expressions. For instance, we can express tanα in terms of cosα, and use the Pythagorean identity to change trigonometric expressions involving sinα to expressions involving cosα alone. This form of equation is particularly useful when dealing with problems that naturally involve angles, such as those found in physics or engineering contexts where direction and orientation are important.