Question

A straight line cuts off the intercepts OA = a and OB = b on the positive directions of the x-axis and y-axis, respectively. If the perpendicular from the origin O to this line makes an angle of \(\frac{\pi}{6}\) with the positive direction of the y-axis and the area of △OAB is \(\frac{98\sqrt{3}}{3},\) then a² − b² is equal to:

• A. \(\frac{392}{3}\)
• B. 196
• C. \(\frac{196}{3}\)
• D. 98

Answer 1

The Correct Option is A

The equation of the straight line is given in intercept form as:

\[ \frac{x}{a} + \frac{y}{b} = 1. \]

Alternatively, the equation of the line in perpendicular form is given as:

\[ x \cos \frac{\pi}{3} + y \sin \frac{\pi}{3} = p. \]

Simplifying this gives:

\[ \frac{x}{2} + \frac{y}{\sqrt{3}/2} = p. \]

Rearranging the terms, we get:

\[ \frac{x}{3p} + \frac{y}{2p} = 1. \]

Comparing the two forms of the equation of the line, we can identify:

• \(a = 2p\)
• \(b = 2p\sqrt{3}\)

Area of \(\triangle OAB\)

The area of \(\triangle OAB\), where \(A\) and \(B\) are the intercepts on the \(x\)- and \(y\)-axes respectively, is given by:

\[ \text{Area} = \frac{1}{2}ab. \]

We are given that this area is \(\frac{98}{\sqrt{3}}\). Substituting the values of \(a\) and \(b\), we have:

\[ \frac{1}{2}(2p)(2p\sqrt{3}) = \frac{98}{\sqrt{3}}. \]

Simplifying, we find:

\[ p^2 = 49. \]

Finding \(a^2 - b^2\)

We are asked to find \(a^2 - b^2\). Using the values of \(a\) and \(b\) in terms of \(p\), we get:

\[ a^2 - b^2 = (2p)^2 - (2p\sqrt{3})^2. \]

Expanding the terms:

\[ a^2 - b^2 = 4p^2 - 4p^2 \cdot 3. \]

Simplify further:

\[ a^2 - b^2 = \frac{8p^2}{3}. \]

Substituting \(p^2 = 49\), we find:

\[ a^2 - b^2 = \frac{8}{3} \cdot 49 = \frac{392}{3}. \]

Conclusion

The value of \(a^2 - b^2\) is:

\[ \boxed{\frac{392}{3}}. \]