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 origin O to this line makes an angle of 6 with the positive direction of the y -axis and the area

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Solution

Q.
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 origin O to this line makes an angle of $\frac{π}{6}$ with the positive direction of the $y$ -axis and the area of $∆ O A B$ is $\frac{98}{3} \sqrt{3},$ then $a^{2} - b^{2}$ is equal to [2023]

1 $\frac{392}{3}$

2 $\frac{196}{3}$

3 $196$

4 $98$

Ans.
(1)

$Let the perpendicular distance be p .$

$The equation of the line A B is given as$

$x \cos \frac{π}{3} + y \sin \frac{π}{3} = p$

$\Rightarrow x \cdot \frac{1}{2} + y \frac{\sqrt{3}}{2} = p$

$\Rightarrow \frac{x}{2} + \frac{y}{(\frac{2}{\sqrt{3}})} = p \Rightarrow \frac{x}{2 p} + \frac{y}{\frac{2 p}{\sqrt{3}}} = 1$ ...(i)

$Intercept form of line A B will be \frac{x}{a} + \frac{y}{b} = 1$ ...(ii)
$Comparing eq (i) and (ii), we get a = 2 p, b = \frac{2 p}{\sqrt{3}}$

$Now, area of triangle O A B = \frac{98}{3} \sqrt{3}$

$S o, \frac{1}{2} a b = \frac{98}{3} \sqrt{3} \Rightarrow \frac{1}{2} \times 2 p \times \frac{2 p}{\sqrt{3}} = \frac{98}{3} \sqrt{3} \Rightarrow 2 p^{2} = 98$

$\Rightarrow p^{2} = 49 \Rightarrow p = 7 (∵ p \geq 0)$

$∴$ $a = 2 \times 7 = 14,$ $b = \frac{2 \times 7}{\sqrt{3}} = \frac{14}{\sqrt{3}}$

$∴$ $a^{2} - b^{2} = {(14)}^{2} - {(\frac{14}{\sqrt{3}})}^{2}$ $= \frac{392}{3}$

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