The locus of the point of intersection of the lines x=a ((1-t^2)/(1+t^2)) and y=(2 a t)/(1+t^2) represent (t being a parameter)

Correct answer: A. Circle Given x=a ((1-t^2)/(1+t^2)) and y=(2 a t)/(1+t^2) aligned & Let t= θ & x=a((1- ^2 θ)/(1+ ^2 θ)) & and y=(2 a θ)/(1+ ^2 θ) & x=a 2 θ and y=a 2 θ & 2 θ=(x)/(a) and 2 θ=(y)/(a) aligned Squaring both sides, we get ^2 2 θ= (x^2)/(a^2) ....(i) and ^2 2 θ= (y^2)/(a^2) ....(ii) Adding Eqs. (i) and (ii), we get aligned ^2 2 θ+ ^2 2 θ& =(x^2)/(a^2)+(y^2)/(a^2) x²+y² & =a² aligned Locus is a circle having centre at origin and radius a .

BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Mathematics
Chapter: Straight Lines

BITSAT 2024 · Mathematics · Straight Lines

The locus of the point of intersection of the lines x=a...

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)

Options

ACircle
BParabola
CEllipse
DHyperbola

Correct answer

A. Circle

Step-by-step 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 } \sin2 \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 x²+y² & =a² \end{aligned}

\therefore

Locus is a circle having centre at origin and radius

a

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