A line which makes an acute angle $\theta$ with the positive direction of $x$-axis is drawn through the point $P(3,4)$ to meet the line $x = 6$ at $R$ and $y = 8$ at $S$, then:
$PR = 3\sec\theta$
$PS = 4\csc\theta$
$PR + PS = \frac{2(3\sin\theta + 4\cos\theta)}{\sin 2\theta}$
$\frac{9}{(PR)^2} + \frac{16}{(PS)^2} = 1$
Step-by-Step Solution
Key Concept: In a rectangle, opposite sides are equal and parallel; use parametric angle relationships to express dimensions.
Given rectangle $ABCD$ with vertices $P(3,4)$, $Q(6,4)$, $S(\beta,8)$, and $R(6,\alpha)$, we use the property that diagonals of a rectangle bisect each other. From $PU = (PR)\cos\theta = \frac{3}{\cos\theta}$, we get $PR = 3\sec\theta$. Similarly, $PT = (PS)\sin\theta = \frac{4}{\sin\theta}$, giving $PS = 4\csc\theta$. The coordinates are determined by the rectangle constraints and the angle $\theta$ that sides make with the axes.
Correct Answer: 1,2,3,4