JEE Main 2002 — Mathematics Coordinate Geometry

Equation of $A B$AB is $x \cos \alpha+y \sin \alpha=p \Rightarrow \frac{x \cos \alpha}{p}+\frac{y \sin \alpha}{p}=1 \Rightarrow \frac{x}{p / \cos \alpha}+\frac{y}{p / \sin \alpha}=1$xcosα+ysinα=p⇒pxcosα​+pysinα​=1⇒p/cosαx​+p/sinαy​=1 So co-ordinates of $A$A and $B$B are $\left(\frac{p}{\cos \alpha}, 0\right)$(cosαp​,0) and $\left(0, \frac{p}{\sin \alpha}\right)$(0,sinαp​); So coordinates of mid point of $A B$AB are $\left(\frac{p}{2 \cos \alpha}, \frac{p}{2 \sin \alpha}\right)=\left(x_1, y_1\right)($(2cosαp​,2sinαp​)=(x1​,y1​)( let $) ; x_1=\frac{p}{2 \cos \alpha} \& y_1=\frac{p}{2 \sin \alpha}$);x1​=2cosαp​&y1​=2sinαp​ $\Rightarrow \cos \alpha=p / 2 x_1$⇒cosα=p/2x1​ and $\sin \alpha=p / 2 y_1 ; \cos ^2 \alpha+\sin ^2 \alpha=1 \Rightarrow \frac{p^2}{4}\left(\frac{1}{x_1^2}+\frac{1}{y_1^2}\right)=1$sinα=p/2y1​;cos2α+sin2α=1⇒4p2​(x12​1​+y12​1​)=1 Locus of $\left(x_1, y_1\right)$(x1​,y1​) is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$x21​+y21​=p24​.