Question

Which of the following equation(s) ($t$ being the parameter) represents a hyperbola?

• A. $\frac{tx}{a}-\frac{y}{b}+t=0,\frac{x}{a}+\frac{ty}{b}-1=0$
• B. $x=\frac{a}{2}(t+\frac{1}{t}),y=\frac{b}{2}(t-\frac{1}{t})$
• C. $x=e^t+e^{-t},y=e^t-e^{-t}$
• D. $x^2=2(\cos t+3),y^2=2({\cos }^2\frac{t}{2}-1)$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $x=\frac{a}{2}(t+\frac{1}{t}),y=\frac{b}{2}(t-\frac{1}{t})$
C $x=e^t+e^{-t},y=e^t-e^{-t}$
D $x^2=2(\cos t+3),y^2=2({\cos }^2\frac{t}{2}-1)$

Option $\left(a\right)$,
$\frac{tx}{a}-\frac{y}{b}+t=0\cdots \left(i\right)$
$\frac{x}{a}+\frac{ty}{b}-1=0\cdots \left(ii\right)$
$\Rightarrow t=\frac{b}{y}(1-\frac{x}{a})$
Put this value in $\left(i\right)$, we get
$\frac{b}{y}(1-\frac{x}{a})(1+\frac{x}{a})-\frac{y}{b}=0$
$\Rightarrow \frac{b}{y}(1-\frac{x^2}{a^2})-\frac{y}{b}=0$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
which represents an ellipse.

Option $\left(b\right)$
$x=\frac{a}{2}(t+\frac{1}{t})$
$\Rightarrow \frac{2x}{a}=t+\frac{1}{t}\cdots \left(iii\right)$
$y=\frac{b}{2}(t-\frac{1}{t})$
$\Rightarrow \frac{2y}{b}=t-\frac{1}{t}\cdots \left(iv\right)$
$\Rightarrow \frac{4x^2}{a^2}-\frac{4y^2}{b^2}=4$
$\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
which represents a hyperbola

Option $\left(c\right)$
$x=e^t+e^{-t}\cdots \left(v\right)$
$y=e^t-e^{-t}\cdots \left(vi\right)$
$\Rightarrow x^2-y^2=4$ which is a hyperbola

Option $\left(d\right)$
$x^2=2(\cos t+3)\cdots \left(vii\right)$
$\Rightarrow x^2=4{\cos }^2\frac{t}{2}+4$
$\Rightarrow \frac{x^2}{4}={\cos }^2\frac{t}{2}+1$
$y^2=2({\cos }^2\frac{t}{2}-1)\cdots \left(viii\right)$
$\Rightarrow \frac{y^2}{2}={\cos }^2\frac{t}{2}-1$
$\Rightarrow \frac{x^2}{8}-\frac{y^2}{4}=1$
which represents a hyperbola.