Show that the point (x, y) given by x= 2at 1+t^2 and y=a ( 1-t^2 1+t^2 )2 lies on a circle for all real values of t such that -1 1, where a is any given real number.

x= 2at 1+t^2 ,y=a ( 1-t^2 1+t ) x^2+y^2= 4a^2t^2 (1+t^2)^2 + a^2(1-t^2)^2 (1+t^2)^2 = 4a^2t^2+a^2(1-2t^2+a^2)t^4 (1+t^2)^2 = 4a^2t^2+a^2-2a^2t^2+a^2t^4 (1+t^2)^2 = 2 a^2t^2+a^2+a^2t^2 (1+t^2)^2 = a^2(1+2t^2+t^2) (1+t^2)^2 x^2+y^2=a^2 is equation of a circle.

Show that the point (x, y) given by x= 2at 1+t^2 and y=a ( 1-t^2 …

Download App

Question

Show that the point $(\text{x},\ \text{y})$ given by $\text{x}=\frac{2\text{at}}{1+\text{t}^2}$ and $\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}^2}\Big)$2 lies on a circle for all real values of t such that $-1\leq\text{t}\leq1,$ where a is any given real number.

✓

Answer

$\text{x}=\frac{2\text{at}}{1+\text{t}^2},\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}}\Big)$
$\text{x}^2+\text{y}^2=\frac{4\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}+\frac{\text{a}^2(1-\text{t}^2)^2}{(1+\text{t}^2)^2}$
$=\frac{4\text{a}^2\text{t}^2+\text{a}^2(1-2\text{t}^2+\text{a}^2)\text{t}^4}{(1+\text{t}^2)^2}$
$=\frac{4\text{a}^2\text{t}^2+\text{a}^2-2\text{a}^2\text{t}^2+\text{a}^2\text{t}^4}{(1+\text{t}^2)^2}$
$=\frac{2\text{a}^2\text{t}^2+\text{a}^2+\text{a}^2\text{t}^2}{(1+\text{t}^2)^2}$
$=\frac{\text{a}^2(1+2\text{t}^2+\text{t}^2)}{(1+\text{t}^2)^2}$
$\text{x}^2+\text{y}^2=\text{a}^2$ is equation of a circle.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from The Circle→The Circle — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Prove the following identities:
$\frac{\cos\text{x}}{1-\sin\text{x}}=\frac{1+\cos\text{x}+\sin\text{x}}{1+\cos\text{x}-\sin\text{x}}$

→

Match the following:

$a.$	If $E_1$ and $E_2$ are the two mutually exclusive events	$i.$	$\text{E}_1\cap\text{E}_2=\text{E}_1$
$b.$	If $E_1$ and $E_2$ are the mutually exclusive and exhaustive events	$ii.$	$(\text{E}_1-\text{E}_2)\cup(\text{E}_1\cap\text{E}_2)=\text{E}_1$
$c.$	If $E_1$ and $E_2$ have common outcomes, then	$iii.$	$\text{E}_1\cap\text{E}_2=\phi,\text{ E}_1\cup\text{E}_2=\text{S}$
$d.$	If $E_1$ and $E_2$ are two events such that $\text{E}_1\subset\text{E}_2$	$iv.$	$\text{E}_1\cap\text{E}_2=\phi$