Question

The equations $x=a\left(\frac{1-t^2}{1+t^2}\right),y=\frac{2bt}{1+t^2};t\in R$ represent

• A. A circle
• B. An ellipse
• C. A parabola
• D. A hyperbola

Answer 1

The correct option is B An ellipse

Consider first equation i.e. $x=a\left(\frac{1-t^2}{1{+t}^2}\right)$

$\therefore \frac{x}{a}=\left(\frac{1-t^2}{1{+t}^2}\right)$

By using componendo- dividendo, we get,

$\therefore \frac{x+a}{x-a}=\left(\frac{(1-t^2)+\left(1{+t}^2\right)}{(1-t^2)-\left(1{+t}^2\right)}\right)$

$\therefore \frac{x+a}{x-a}=\left(\frac{2}{-2t^2}\right)$

$\therefore \frac{x+a}{x-a}=\frac{-1}{t^2}$

Taking reciprocals on both sides, we get,

$\therefore -t^2=\frac{x-a}{x+a}$

Multiplying both sides by -1, we get,

$\therefore t^2=\frac{-(x-a)}{x+a}$

$\therefore t^2=\frac{a-x}{x+a}$ Equation (1)

Now consider second equation i.e. $y=\frac{2bt}{1+t^2}$

Squaring both sides, we get,

$\therefore y^2=\frac{4b^2{t}^2}{{(1+t^2)}^2}$

Put value of $t^2$ from equation (1),

$\therefore y^2=\frac{4b^2\left(\frac{a-x}{a+x}\right)}{{(1+\frac{a-x}{a+x})}^2}$

$\therefore y^2=\frac{4b^2\left(\frac{a-x}{a+x}\right)}{{\left(\frac{(a+x)+(a-x)}{a+x}\right)}^2}$

$\therefore y^2=\frac{4b^2\left(\frac{a-x}{a+x}\right)}{{\left(\frac{2a}{a+x}\right)}^2}$

$\therefore y^2=\frac{4b^2\left(\frac{a-x}{a+x}\right)}{\frac{4a^2}{{(a+x)}^2}}$

$\therefore y^2=b^2\left(\frac{a-x}{a+x}\right)\times \frac{{(a+x)}^2}{a^2}$

$\therefore y^2=\frac{b^2(a-x)(a+x)}{a^2}$

$\therefore a^2{y}^2=b^2(a-x)(a+x)$

$\therefore a^2{y}^2=b^2(a^2-x^2)$

$\therefore a^2{y}^2=a^2{b}^2-x^2{b}^2$

$\therefore x^2{b}^2+a^2{y}^2=a^2{b}^2$

Dividing both sides by $a^2{b}^2$, we get,

$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\therefore {\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2=1$

This is equation of an ellipse.

Thus, answer is option (B)