Question

If lines be drawn parallel to the axes of co-ordinates from the points where x cosα + y sinα = p meets them so as to meet the perpendicular on this line from the origin in the points P and Q then prove that | PQ | = 4p | cos2α | cosec^2(2α.)

Answer 1

Answer the question #56866 – Math – Analytic Geometry

If lines be drawn parallel to the axes of coordinates from the points where $x \cos \alpha + y \sin \alpha = p$ meets them so as to meet the perpendicular on this line from the origin in the points $P$ and $Q$ then prove that $|PQ| = 4p|\cos 2\alpha| \cdot \csc^2(2\alpha)$.

Solution

Let $AB$ is the line, which shows by the given equation $x \cos \alpha + y \sin \alpha = p$ (see Fig. below). Then coordinates of the points $A$ and $B$ are $\left(0; \frac{p}{\sin \alpha}\right)$ and $\left(\frac{p}{\cos \alpha}; 0\right)$ accordingly. After that we write the equation for the line $PQ$ which is perpendicular to $AB$ and crosses the origin $(0; 0)$. We obtain $PQ: x \sin \alpha - y \cos \alpha = 0$.

Equation of line $AD$ is $y = \frac{p}{\sin \alpha}$, hence the y-coordinate of point $P$ is $y_P = \frac{p}{\sin \alpha}$.

Equation of line $BD$ is $x = \frac{p}{\cos \alpha}$, hence the x-coordinate of point $Q$ is $x_p = \frac{p}{\cos \alpha}$. Coordinates of the points $P$ and $Q$ satisfy equation $PQ$: $x \sin \alpha - y \cos \alpha = 0$, so they are $\left(\frac{p \cos \alpha}{\sin^2 \alpha}; \frac{p}{\sin \alpha}\right)$ and $\left(\frac{p}{\cos \alpha}; \frac{p \sin \alpha}{\cos^2 \alpha}\right)$ respectively. By the formula $|PQ| = \sqrt{\left(x_q - x_p\right)^2 + \left(y_q - y_p\right)^2}$, we obtain:

$$\begin{array}{l} |PQ| = \sqrt{\left(\frac{p}{\cos \alpha} - \frac{p \cos \alpha}{\sin^2 \alpha}\right)^2 + \left(\frac{p \sin \alpha}{\cos^2 \alpha} - \frac{p}{\sin \alpha}\right)^2} \\ = p \sqrt{\left(\frac{\sin^2 \alpha - \cos^2 \alpha}{\cos \alpha \sin^2 \alpha}\right)^2 + \left(\frac{\sin^2 \alpha - \cos^2 \alpha}{\cos^2 \alpha \sin \alpha}\right)^2} = p \sqrt{\frac{\cos^2 (2\alpha)}{\cos^2 \alpha \sin^4 \alpha} + \frac{\cos^2 (2\alpha)}{\cos^4 \alpha \sin^2 \alpha}} \\ = \frac{4p |\cos (2\alpha)|}{\sqrt{16 \cos^4 \alpha \sin^4 \alpha}} = \frac{4p |\cos (2\alpha)|}{\sqrt{(\sin (2\alpha))^4}} = 4p |\cos (2\alpha)| \cdot \csc^2 (2\alpha) \end{array}$$

Thus, $|PQ| = 4p |\cos 2\alpha| \cdot \csc^2(2\alpha)$.

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