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Chapter 15: Problem 1460

If PN is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, of the midpoint of PN is (a) A circle (b) a hyperbola (c) a parabola (d) An ellipse

Short Answer

Expert verified
In summary, the locus of the midpoint of PN is an ellipse, given by the equation \(\frac{(hk)^2}{c^2} - \frac{h^2+k^2}{1} + c^2 = 0\), where P is a point on the rectangular hyperbola \(xy=c^2\), N is the foot of the perpendicular from P to its asymptote, and M is the midpoint of the segment PN. The correct option is (d) An ellipse.

Step by step solution

01

Rectangular Hyperbola Equation

Let the rectangular hyperbola equation be \(xy = c^2\), where \(c\) is a constant. The asymptotes of this hyperbola are given by the equations \(y = \pm \frac{c^2}{x}\).
02

Point on Hyperbola and Coordinates

Let P be the point \((x_1, y_1)\) on the hyperbola, such that \(x_1y_1 =c^2\). Let N be the foot of the perpendicular from P on the asymptote.
03

Foot of the Perpendicular onto the Asymptote and Coordinates

The coordinates of N will be \(\left(\frac{c^2}{x_1}, y_1\right)\) or \(\left(x_1, \frac{c^2}{y_1}\right)\), depending on which asymptote the perpendicular segment is drawn.
04

Midpoint Coordinates of PN

Let M be the midpoint of the segment PN. We can compute the coordinates of M with the average of the coordinates: M=\(\left(\frac{x_1+\frac{c^2}{x_1}}{2}, \frac{y_1+\frac{c^2}{y_1}}{2}\right)\)
05

Replacement

Since \(x_1y_1 = c^2\), we can replace in the coordinates of M: M=\(\left(\frac{x_1^2+c^2}{2x_1}, \frac{y_1^2+c^2}{2y_1}\right)\)
06

Locus Equation

Let's now consider the locus h and k of the point M, so we get: h = \(\frac{x_1^2+c^2}{2x_1}\) and k = \(\frac{y_1^2+c^2}{2y_1}\) Now, we can express x1 and y1 in terms of h and k: \(x_1^2 =2hx_1 -c^2\) and \(y_1^2 = 2ky_1 -c^2\) from which we obtain: \(x_1 = h \pm \sqrt{h^2 -c^2}\) and \(y_1 = k \pm \sqrt{k^2 -c^2}\)
07

Rectangular Hyperbola Condition

We know that the point P \((x_1, y_1)\) lies on the rectangular hyperbola, so its coordinates satisfy \(x_1y_1 =c^2\). We can use this condition to eliminate either x1 or y1: \((h \pm \sqrt{h^2 -c^2})(k \pm \sqrt{k^2 -c^2}) = c^2\)
08

Simplification and Locus Equation

After some algebraic manipulations, we obtain: \((hk)^2 - c^2(h^2 + k^2) + c^4 = 0\) This is the equation of the locus of the midpoint M of the segment PN.
09

Identifying the Locus Shape

We can identify the type of the locus of the midpoint by determining whether it is a circle, hyperbola, ellipse, or parabola. If you look at the locus equation, it looks like a quadratic function and can be rearranged into a form that will help us identify its type: \((hk)^2 - c^2(h^2 + k^2) + c^4 = 0\) \(\Rightarrow \frac{(hk)^2}{c^2} - \frac{h^2+k^2}{1} + c^2 = 0\) After rearranging the equation in this form, we can notice that the equation represents an ellipse. Therefore, the locus of the midpoint of PN is an ellipse. So, the correct option is (d) An ellipse.

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