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:
\(\frac{392}{3}\)
196
\(\frac{196}{3}\)
98
The Correct Option is A
Solution and Explanation
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:
\[ \boxed{\frac{392}{3}}. \]
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