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QJEEM23Ma9618: JEE Main 2023 Mathematics - Class 11 JEEM23Ma9618 Solution: A straight line cuts off the intercepts and on the positive…

A straight line cuts off the intercepts OA=a\mathrm{OA}=\mathrm{a} and OB=b\mathrm{OB}=\mathrm{b} on the positive directions of xx -axis and yy axis respectively. If the perpendicular from origin OO to this line makes an angle of π6\frac{\pi}{6} with positive direction of yy -axis and the area of △OAB\triangle \mathrm{OAB} is 9833\frac{98}{3} \sqrt{3} , then a2−b2\mathrm{a}^{2}-\mathrm{b}^{2} is equal to :

Multiple Choice Options

  • A 3923\frac{392}{3}
  • B 98
  • C 196
  • D 1963\frac{196}{3}

Step-by-Step Verified Solution

12ab=9833{1 \over 2}ab = {{98\sqrt 3 } \over 3}

JEE Main 2023 (Online) 30th January Morning Shift Mathematics - Properties of Triangle Question 10 English Explanation

⇒3ab=196\Rightarrow \sqrt 3 ab = 196 ..... (i)

OP=OBcos⁡30∘=OAcos⁡60∘OP = OB\cos 30^\circ = OA\cos 60^\circ

⇒b32=a2\Rightarrow {{b\sqrt 3 } \over 2} = {a \over 2}

⇒3b=a\Rightarrow \sqrt 3 b = a ..... (ii)

By (i) and (ii)

a2=196{a^2} = 196

a=14a = 14

b2=a23{b^2} = {{{a^2}} \over 3}

a2−b2=2a23=3923{a^2} - {b^2} = {{2{a^2}} \over 3} = {{392} \over 3}

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