JEE Main 2023 — Mathematics Coordinate Geometry

Plotting the diagram of the given data we get,

Equation of straight line

Here given OA=a&OB=b, so equation of line $AB$AB in intercept form will be $\frac{x}{a}+\frac{y}{b}=1......(1)$ax​+by​=1......(1)

And given perpendicular line of length $p$p is making $\frac{\pi }{3}$3π​ angle with the $x-$x−axis,

So, the equation of line in normal form will be

$$x\mathrm{cos}(\frac{\pi }{3})+y\mathrm{sin}(\frac{\pi }{3})=p$$xcos(3π​)+ysin(3π​)=p

$$\Rightarrow \frac{x}{2p}+\frac{y\sqrt{3}}{2p}=1$$⇒2px​+2py3​​=1

So, on comparing with the equation $(1)$(1) we get,

a=2p&b=\frac{2p}{\sqrt{3}}

Now area of the triangle $OAB$OAB is

$$\frac{1}{2}\times a\times b=\frac{98\sqrt{3}}{3}$$21​×a×b=3983​​

$$\Rightarrow \frac{1}{2}\times 2p\times \frac{2p}{\sqrt{3}}=\frac{98\sqrt{3}}{3}$$⇒21​×2p×3​2p​=3983​​

$$\Rightarrow {p}^{2}=49$$⇒p2=49

Therefore,

$${a}^{2}-{b}^{2}=4{p}^{2}-\frac{4{p}^{2}}{3}=\frac{8}{3}{p}^{2}$$a2−b2=4p2−34p2​=38​p2

$$\Rightarrow {a}^{2}-{b}^{2}=\frac{8}{3}(49)=\frac{392}{3}$$⇒a2−b2=38​(49)=3392​