JEE Main 2021 — Mathematics Coordinate Geometry

$${L}_{1}:xcosec\alpha -y\mathrm{sec}\alpha =k\mathrm{cot}2\alpha$$L1​:xcosecα−ysecα=kcot2α

$$\frac{x}{\mathrm{sin}\alpha }-\frac{y}{\mathrm{cos}\alpha }=k\frac{\mathrm{cos}2\alpha }{\mathrm{sin}2\alpha }$$sinαx​−cosαy​=ksin2αcos2α​

$$\frac{x\mathrm{cos}\alpha -y\mathrm{sin}\alpha }{\mathrm{sin}\alpha \mathrm{cos}\alpha }=k\frac{\mathrm{cos}2\alpha }{2\mathrm{sin}\alpha \mathrm{cos}\alpha }$$sinαcosαxcosα−ysinα​=k2sinαcosαcos2α​

$$\Rightarrow x\mathrm{cos}\alpha -y\mathrm{sin}\alpha =\frac{k}{2}\mathrm{cos}2\alpha$$⇒xcosα−ysinα=2k​cos2α

We know that Perpendicular distance from a point $({x}_{1},{y}_{1})$(x1​,y1​)to $ax+by+c=0$ax+by+c=0 is $|\frac{a{x}_{1}+b{y}_{1}+c}{\sqrt{{a}^{2}+{b}^{2}}}|$∣a2+b2​ax1​+by1​+c​∣

Perpendicular distance from $(0,0)$(0,0) is $p=|\frac{0-0-\frac{k}{2}\mathrm{cos}2\alpha }{\sqrt{{\mathrm{cos}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha }}|$p=∣cos2α+sin2α​0−0−2k​cos2α​∣

$$\Rightarrow {p}^{2}=\frac{{k}^{2}}{4}{\mathrm{cos}}^{2}2\alpha$$⇒p2=4k2​cos22α

$${L}_{2}:x\mathrm{sin}\alpha +y\mathrm{cos}\alpha -k\mathrm{sin}2\alpha =0$$L2​:xsinα+ycosα−ksin2α=0

Perpendicular distance from $(0,0)$(0,0) is $q=|\frac{0+0-k\mathrm{sin}2\alpha }{\sqrt{{\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha }}|$q=∣sin2α+cos2α​0+0−ksin2α​∣

$$\Rightarrow {q}^{2}={k}^{2}{\mathrm{sin}}^{2}2\alpha$$⇒q2=k2sin22α

$$4{p}^{2}+{q}^{2}={k}^{2}({\mathrm{sin}}^{2}2\alpha +{\mathrm{cos}}^{2}2\alpha )$$4p2+q2=k2(sin22α+cos22α)

We know that ${\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1.$sin2t+cos2t=1.

$$4{p}^{2}+{q}^{2}={k}^{2}$$4p2+q2=k2