JEE Main 2019 — Mathematics Coordinate Geometry

Any point on the circle ${x}^{2}+{y}^{2}={r}^{2}$x2+y2=r2 is $(r\mathrm{cos}\theta ,r\mathrm{sin}\theta )$(rcosθ,rsinθ) and the equation of the tangent at this point is $x\mathrm{cos}\theta +y\mathrm{sin}\theta =r.$xcosθ+ysinθ=r.

Thus, any point on circle ${x}^{2}+{y}^{2}=1$x2+y2=1 is $(\mathrm{cos}\theta ,\mathrm{sin}\theta )$(cosθ,sinθ) and the equation of tangent at this point is $x\mathrm{cos}\theta +y\mathrm{sin}\theta =1.$xcosθ+ysinθ=1.

To find the point where this line intersects the $x$x-axis, put $y=0,$y=0,$\Rightarrow x=\frac{1}{\mathrm{cos}\theta }$⇒x=cosθ1​

$$\Rightarrow P=(\frac{1}{\mathrm{cos}\theta },0)$$⇒P=(cosθ1​,0)

And, to find the point where this line intersects the $y$y-axis, put $x=0,$x=0,

$$\Rightarrow y=\frac{1}{\mathrm{sin}\theta }$$⇒y=sinθ1​

$$\Rightarrow Q=(0,\frac{1}{\mathrm{sin}\theta })$$⇒Q=(0,sinθ1​)

The mid-point of a line segment joining the points $({x}_{1},{y}_{1})$(x1​,y1​) and $({x}_{2},{y}_{2})$(x2​,y2​) is $(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})$(2x1​+x2​​,2y1​+y2​​)

Let, midpoint of $PQ$PQ is $(h,k)$(h,k)

$$\therefore (h,k)=(\frac{1}{2\mathrm{cos}\theta },\frac{1}{2\mathrm{sin}\theta })$$∴(h,k)=(2cosθ1​,2sinθ1​)

\Rightarrow h=\frac{1}{2\mathrm{cos}\theta } & k=\frac{1}{2\mathrm{sin}\theta }

\Rightarrow \mathrm{cos}\theta =\frac{1}{2h} & \mathrm{sin}\theta =\frac{1}{2k}

$$\because {\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$$∵sin2θ+cos2θ=1

$$\Rightarrow \frac{1}{4{h}^{2}}+\frac{1}{4{k}^{2}}=1$$⇒4h21​+4k21​=1

$$\Rightarrow \frac{1}{{h}^{2}}+\frac{1}{{k}^{2}}=4$$⇒h21​+k21​=4

To get the locus of the required point, replace $(h,k)$(h,k) by $(x,y)$(x,y)

$\therefore$∴Locus is $\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}=4$x21​+y21​=4

$$\Rightarrow {x}^{2}+{y}^{2}–4{x}^{2}{y}^{2}=0.$$⇒x2+y2–4x2y2=0.