Question

If t is a non-zero parameter then the point lies on$\left(\frac{a}{2}(t+\frac{1}{t})\right)$, $\left(\frac{b}{2}(t-\frac{1}{t})\right)$ lies on

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

Answer 1

The correct option is D hyperbola

Given point t:$\left(\frac{a}{2}\right(t+\frac{1}{t}\left)\right),\left(\frac{b}{2}\right(t-\frac{1}{t}\left)\right)$

A: Circle $x^2+y^{}=r^2=(\sqrt{a^2+b^2}{)}^2$

$\Rightarrow \left[\frac{a}{2}\right(t+\frac{1}{t}){]}^2+[\frac{b}{2}(t-\frac{1}{t}){]}^2=a^2+b^2$

$\Rightarrow \frac{a^2}{4}(t+\frac{1}{t}{)}^2+\frac{b^2}{4}(t-\frac{1}{t}{)}^2=a^2+b^2$

We see, we cannot eliminate 't' from it so it does not satisfy.

B: Parabola: No 'b' exists there generally.

C: Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{1}{a^2}\cdot \left[\frac{a}{2}\right(t+\frac{1}{t}){]}^2+\frac{1}{b^2}[\frac{b}{2}(t-\frac{1}{t}){]}^2=1$

$\Rightarrow \frac{1}{4}(t+\frac{1}{t}{)}^2+\frac{1}{4}(t-\frac{1}{t}{)}^2=1$

$\Rightarrow t^2+\frac{1}{t^2}+2+t^2+\frac{1}{t^2}-2=4$

$\Rightarrow 2t^2+\frac{2}{t^2}\ne 4$ (Not an ellipse)

D: Hyperbola :$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\Rightarrow \frac{1}{a^2}\cdot \left[\frac{a}{2}\right(t+\frac{1}{t}){]}^2-\frac{1}{b^2}[\frac{b}{2}(t-\frac{1}{t}){]}^2=1$

$\Rightarrow \frac{1}{4}(t+\frac{1}{t}{)}^2-\frac{1}{4}(t-\frac{1}{t}{)}^2=1$

$\Rightarrow t^2+\frac{1}{t^2}+2-(t^2-\frac{1}{t^2}+2)=4$

$\Rightarrow 2+2=4$

So, point lies on Hyperbola.