JEE Challenger
More from Inverse Trigonometric Functions
MathHardAdv 2007 • Paper 2

Locus of Points Governed by Inverse Trigonometric Equation

Let (x,y)(x, y) be such that
sin⁡−1(ax)+cos⁡−1(y)+cos⁡−1(bxy)=π2.\sin^{-1}(ax) + \cos^{-1}(y) + \cos^{-1}(bxy) = \frac{\pi}{2}.

Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the 4×44 \times 4 matrix given in the ORS.

Column IColumn II(A) If a=1 and b=0, then (x,y)(p) lies on the circle x2+y2=1(B) If a=1 and b=1, then (x,y)(q) lies on (x2−1)(y2−1)=0(C) If a=1 and b=2, then (x,y)(r) lies on y=x(D) If a=2 and b=2, then (x,y)(s) lies on (4x2−1)(y2−1)=0\begin{array}{ll} \textbf{Column I} & \textbf{Column II} \\[6pt] \text{(A) If } a = 1 \text{ and } b = 0\text{, then } (x, y) & \text{(p) lies on the circle } x^2 + y^2 = 1 \\[6pt] \text{(B) If } a = 1 \text{ and } b = 1\text{, then } (x, y) & \text{(q) lies on } (x^2 - 1)(y^2 - 1) = 0 \\[6pt] \text{(C) If } a = 1 \text{ and } b = 2\text{, then } (x, y) & \text{(r) lies on } y = x \\[6pt] \text{(D) If } a = 2 \text{ and } b = 2\text{, then } (x, y) & \text{(s) lies on } (4x^2 - 1)(y^2 - 1) = 0 \end{array}

Options

A

(A)→(p),  (B)→(r),  (C)→(p),  (D)→(s)(\text{A}) \rightarrow (\text{p}),\; (\text{B}) \rightarrow (\text{r}),\; (\text{C}) \rightarrow (\text{p}),\; (\text{D}) \rightarrow (\text{s})

B

(A)→(q),  (B)→(p),  (C)→(s),  (D)→(p)(\text{A}) \rightarrow (\text{q}),\; (\text{B}) \rightarrow (\text{p}),\; (\text{C}) \rightarrow (\text{s}),\; (\text{D}) \rightarrow (\text{p})

C

(A)→(p),  (B)→(q),  (C)→(p),  (D)→(s)(\text{A}) \rightarrow (\text{p}),\; (\text{B}) \rightarrow (\text{q}),\; (\text{C}) \rightarrow (\text{p}),\; (\text{D}) \rightarrow (\text{s})

Correct
D

(A)→(s),  (B)→(q),  (C)→(p),  (D)→(r)(\text{A}) \rightarrow (\text{s}),\; (\text{B}) \rightarrow (\text{q}),\; (\text{C}) \rightarrow (\text{p}),\; (\text{D}) \rightarrow (\text{r})

Topics & Concepts

Inverse Trigonometric FunctionsInverse Trigonometric EquationsLocusCircle

Step-by-Step Solution

Using the identity sin⁡−1(ax)=π2−cos⁡−1(ax)\sin^{-1}(ax) = \frac{\pi}{2} - \cos^{-1}(ax), the given relation becomes:
π2−cos⁡−1(ax)+cos⁡−1(y)+cos⁡−1(bxy)=π2  ⟹  cos⁡−1(y)+cos⁡−1(bxy)=cos⁡−1(ax)\frac{\pi}{2} - \cos^{-1}(ax) + \cos^{-1}(y) + \cos^{-1}(bxy) = \frac{\pi}{2} \implies \cos^{-1}(y) + \cos^{-1}(bxy) = \cos^{-1}(ax)

Taking cosines of both sides using cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A \cos B - \sin A \sin B:
y(bxy)−1−y21−b2x2y2=ax  ⟹  bxy2−ax=1−y21−b2x2y2y(bxy) - \sqrt{1 - y^2}\sqrt{1 - b^2x^2y^2} = ax \implies bxy^2 - ax = \sqrt{1 - y^2}\sqrt{1 - b^2x^2y^2}
Squaring both sides:
x2(by2−a)2=(1−y2)(1−b2x2y2)x^2(by^2 - a)^2 = (1 - y^2)(1 - b^2 x^2 y^2)
x2(b2y4−2aby2+a2)=1−y2−b2x2y2+b2x2y4x^2(b^2y^4 - 2aby^2 + a^2) = 1 - y^2 - b^2 x^2 y^2 + b^2 x^2 y^4
Cancelling b2x2y4b^2 x^2 y^4:
a2x2−2abx2y2=1−y2−b2x2y2a^2 x^2 - 2abx^2y^2 = 1 - y^2 - b^2x^2y^2
a2x2+y2+(b2−2ab)x2y2=1a^2 x^2 + y^2 + (b^2 - 2ab)x^2y^2 = 1

Evaluating the conditions:

  1. (A) a=1,b=0a = 1, b = 0:
    x2+y2+0=1  ⟹  x2+y2=1(p)x^2 + y^2 + 0 = 1 \implies x^2 + y^2 = 1 \quad (p)

  2. (B) a=1,b=1a = 1, b = 1:
    x2+y2+(1−2)x2y2=1  ⟹  x2+y2−x2y2−1=0  ⟹  (x2−1)(1−y2)=0  ⟹  (x2−1)(y2−1)=0(q)x^2 + y^2 + (1 - 2)x^2y^2 = 1 \implies x^2 + y^2 - x^2y^2 - 1 = 0 \implies (x^2 - 1)(1 - y^2) = 0 \implies (x^2 - 1)(y^2 - 1) = 0 \quad (q)

  3. (C) a=1,b=2a = 1, b = 2:
    x2+y2+(4−4)x2y2=1  ⟹  x2+y2=1(p)x^2 + y^2 + (4 - 4)x^2y^2 = 1 \implies x^2 + y^2 = 1 \quad (p)

  4. (D) a=2,b=2a = 2, b = 2:
    4x2+y2+(4−8)x2y2=1  ⟹  4x2+y2−4x2y2−1=0  ⟹  (4x2−1)(1−y2)=0  ⟹  (4x2−1)(y2−1)=0(s)4x^2 + y^2 + (4 - 8)x^2y^2 = 1 \implies 4x^2 + y^2 - 4x^2y^2 - 1 = 0 \implies (4x^2 - 1)(1 - y^2) = 0 \implies (4x^2 - 1)(y^2 - 1) = 0 \quad (s)

Matching:
(A)→(p),  (B)→(q),  (C)→(p),  (D)→(s)(\text{A}) \rightarrow (\text{p}),\; (\text{B}) \rightarrow (\text{q}),\; (\text{C}) \rightarrow (\text{p}),\; (\text{D}) \rightarrow (\text{s})

Correct Answer: C

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