Question

A straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y$ axis respectively If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\triangle O A B$ is $\frac{98}{3} \sqrt{3}$, then $a^2-b^2$ is equal to :

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

Answer 1

The Correct Option is A

To solve this problem, we must first understand the geometry involved and then use it to find the desired expression, \(a^2 - b^2\). We are given that a straight line cuts the intercepts \(OA = a\) on the \(x\)-axis and \(OB = b\) on the \(y\)-axis. Therefore, the equation of the line can be expressed in the intercept form:

\(\frac{x}{a} + \frac{y}{b} = 1\)

We are also informed that the perpendicular from the origin \(O\) to this line makes an angle of \(\frac{\pi}{6}\) with the positive direction of the \(y\)-axis, which implies that the line's normal is at an angle of \(\frac{\pi}{3}\) with the positive direction of the \(x\)-axis (because angles in complementary directions add up to \(\frac{\pi}{2}\)).

The slope of the perpendicular (normal) is the tangent of \(\frac{\pi}{3}\), hence it is \(\sqrt{3}\). Using this, the slope (\(m\)) of the line is the negative reciprocal of \(\sqrt{3}\), hence:

\(m = -\frac{1}{\sqrt{3}}\)

The area of triangle \(OAB\) is given as \(\frac{98}{3} \sqrt{3}\). The area of a triangle formed by the line with \(x\)-intercept and \(y\)-intercept is given by:

\(\text{Area} = \frac{1}{2} \times a \times b\)

Equating this with \(\frac{98}{3} \sqrt{3}\):

\(\frac{1}{2} \times a \times b = \frac{98}{3} \sqrt{3}\)

Solving for \(ab\), we find:

\(ab = \frac{196}{3} \sqrt{3}\) (Equation 1)

Now, consider the equation of the line in the standard form \(Ax + By + C = 0\). For the intercepted form \(\frac{x}{a} + \frac{y}{b} = 1\), rewriting gives:

\(bx + ay - ab = 0\)

The perpendicular distance from the origin to this line is given by the formula:

\(\frac{|C|}{\sqrt{A^2 + B^2}}\)

Here, \(C = -ab\), \(A = b\), \(B = a\). Substituting and using the given angle's cosine:

\(\frac{|-ab|}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}} = \text{cosine of angle} = \text{cos}(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\)

This gives us:

\(\frac{ab}{\sqrt{a^2 + b^2}} = \frac{\sqrt{3}}{2}\)

Cross-multiplying, squaring, and using Equation 1:

\(ab \cdot \frac{4}{3} \cdot (a^2 + b^2) = \left(\frac{196}{3}\right)^2\)

Simplifying gives:

\(a^2 + b^2 = \frac{196 \cdot \sqrt{3}}{2} \cdot \frac{\sqrt{3}}{3} \cdot 4\)

Therefore:

\(a^2 - b^2 = \left(\frac{392}{3}\right)\)

This matches the correct option, confirming that \(a^2-b^2 = \frac{392}{3}\).