Question

Let \(A\left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)\), \(a \gt 0\), be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If \(D(3\cos\theta, a\sin\theta)\) is a point in the fourth quadrant such that the maximum area of \(\Delta ACD\) is 12 square units, then a is equal to ____________.

Hint:

B = (-3/sqrt(a), sqrt(a)) and C = (-3/sqrt(a), -sqrt(a)), so AC passes through the origin; the area becomes \(3\sqrt{a}(\cos\theta - \sin\theta)\) with maximum \(3\sqrt{2a}\).

Answer 1

Correct Answer: 8

Image of A in the y-axis: \(B\left(-\frac{3}{\sqrt a}, \sqrt a\right)\). Image of B in the x-axis: \(C\left(-\frac{3}{\sqrt a}, -\sqrt a\right)\). So A and C are symmetric about the origin and the line AC passes through the origin with slope \(\frac{a}{3}\): \(ax - 3y = 0\).
\(AC = 2\sqrt{\frac{9}{a} + a} = 2\sqrt{\frac{a^2 + 9}{a}}\).
Distance of \(D(3\cos\theta, a\sin\theta)\) from AC: \(\frac{|3a\cos\theta - 3a\sin\theta|}{\sqrt{a^2 + 9}}\).
Area \(= \frac12 \cdot 2\sqrt{\frac{a^2+9}{a}} \cdot \frac{3a|\cos\theta - \sin\theta|}{\sqrt{a^2+9}} = 3\sqrt{a}\,|\cos\theta - \sin\theta|\).
In the fourth quadrant \(\cos\theta \gt 0 \gt \sin\theta\), so \(\cos\theta - \sin\theta \gt 0\) and its maximum is \(\sqrt2\) (at \(\theta = -\frac{\pi}{4}\), which lies in the fourth quadrant).
Maximum area \(= 3\sqrt{2a} = 12 \Rightarrow \sqrt{2a} = 4 \Rightarrow 2a = 16 \Rightarrow a = 8\).
Answer: 8