Q35 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics
The locus of the point of intersection of the lines $x=a\left(\frac{1-t^2}{1+t^2}\right)$ and $y=\frac{2 a t}{1+t^2}$ represent $(t$ being a parameter)
- Correct answerACircle
- BParabola
- CEllipse
- DHyperbola
Solution
Given $x=a\left(\frac{1-t^2}{1+t^2}\right)$ and $y=\frac{2 a t}{1+t^2}$
$\begin{aligned}
& \text { Let } \quad t=\tan \theta \\
& \Rightarrow \quad x=a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\
& \text { and } \quad y=\frac{2 a \tan \theta}{1+\tan ^2 \theta} \\
& \Rightarrow \quad x=a \cos 2 \theta \text { and } y=a \sin 2 \theta \\
& \Rightarrow \cos 2 \theta=\frac{x}{a} \text { and } \sin 2 \theta=\frac{y}{a}
\end{aligned}$
Squaring both sides, we get
$\cos ^2 2 \theta=\frac{x^2}{a^2}$....(i)
and
$\sin ^2 2 \theta=\frac{y^2}{a^2}$....(ii)
Adding Eqs. (i) and (ii), we get
$\begin{aligned}
\quad \cos ^2 2 \theta+\sin ^2 2 \theta & =\frac{x^2}{a^2}+\frac{y^2}{a^2} \\
\Rightarrow \quad x^2+y^2 & =a^2
\end{aligned}$
$\therefore$ Locus is a circle having centre at origin and radius $a$.
Hence, the correct answer is option: 'A'.