Let A(3/√a,√a), a>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 θ,a sin θ) is a point in the fourth quadrant such that the maximum area of ΔACD is 12 square units, then a is equal to ____________.

Solution

Maths sir · 4 steps

C turns out to be the reflection of A in the origin, so the area becomes one neat expression in θ.

1. Find B and C. Reflecting in the y-axis flips the x-sign, so B = (-3/√a, √a). Reflecting that in the x-axis flips the y-sign, so C = (-3/√a, -√a). That is exactly -A, so the origin O is the midpoint of AC.
2. Area through O. Since O is the midpoint of AC, triangle ACD has twice the area of triangle OAD. The area of OAD is 1/2|x_A y_D - y_A x_D|, so the factor 1/2 cancels.

   Area = |3/√a · a sin θ - √a · 3 cos θ| = 3√a(cos θ - sin θ)

3. Maximise over θ. In the fourth quadrant cos θ > 0 and sin θ < 0, so cos θ - sin θ is positive. It equals √2 cos(θ + π/4), which reaches its maximum √2 at θ = -π/4, a fourth-quadrant angle.
4. So. Maximum area = 3√2√a = 12, so √a = 2√2 and a = 8.

The common slip: 32. Keeping the 1/2 from the triangle formula on top of |x_A y_D - y_A x_D|, without doubling for the two halves of AC, gives 3√2a/2 = 12 and so a = 32.

The idea to keep. When two vertices are reflections of each other in the origin, the area is |x₁y₂ - x₂y₁| using one of them and the third vertex. Then write p cos θ + q sin θ as a single cosine to find its maximum.

Answer from the NTA final answer key. Solution by PrepWiser.