Correct choice is (b) 2

Explanation: The equation of characteristic curve is given by

\(\frac {dy}{dx_{char}} = \frac {- \frac {uv}{a^{2}} ± \sqrt {(-ve \, term)}}{1 – \frac {u^{2}}{a^{2}}}\)

Where u and v are the x and y – component of velocity V. u = Vcosθ and v = Vsinθ. On substituting these values we GET

\(\frac {dy}{dx_{char}} = \frac {- \frac {V^2 cosθsinθ}{a^{2}} ± \sqrt {\frac {V^{2}}{a^{2}} (cos^2 θ + sin^2 θ) – 1}}{1 – \frac {V^{2}}{a^{2}} cos^2 θ}\)

The Mach angle is given by μ = sin^-1\(\frac {1}{M}\) which can be rearranged as follows sinμ = \(\frac {1}{M}\). But Mach NUMBER is given by M^2 = \( \frac {V^{2}}{a^{2}} = \frac {1}{sin^2 μ}\). Substituting these we get,

\(\frac {dy}{dx_{char}} = \frac {- \frac {cosθsinθ}{sin^2 μ}±\sqrt {\frac {(cos^2 θ + sin^2 θ)}{sin^2 μ} – 1}}{1 – \frac {cos^2 θ}{sin^2 μ}}\)

USING trigonometry relations, cos^2 θ + sin^2 θ = 1, THUS the square root term in the numerator becomes

\( \sqrt {\frac {(cos^2 θ + sin^2 θ)}{sin^2 μ} – 1} = \sqrt {\frac {1}{sin^2 μ} – 1} = \sqrt {cosec^2 μ – 1} = \sqrt {cot^2 μ – 1} = \frac {1}{tanμ} \)

The characteristic curve thus becomes

\(\frac {dy}{dx_{char}} = \frac {- \frac {cosθsinθ}{sin^2 μ}±\frac {1}{tanμ}}{1 – \frac {cos^2 θ}{sin^2 μ}}\) = tan⁡(θ ± μ)

This results in two characteristic lines through a point in the flow field. One with an angle tan⁡(θ + μ) and the other tan⁡(θ – μ).