Question

Which of the following equations ($t$ being the parameter) can't represent 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. $x^2=2(\cos t+3)$ , $y^2=2\left(\begin{array}{c}co{s}^2\frac{t}{2}-1\end{array}\right)$

Answer 1

Answer: (A)

$\frac{tx}{a}-\frac{y}{b}+t=0,\frac{x}{a}+\frac{ty}{b}-1=0$

option A,

$\frac{tx}{a}-\frac{y}{b}+t=0$ .....(i)

$\frac{x}{a}+\frac{ty}{b}-1=0$ .....(ii)

$\Rightarrow t=\frac{b}{y}(1-\frac{x}{a})$

Put this value in (i), we get

$\frac{b}{y}(1-\frac{x}{a})(1+\frac{x}{a})-\frac{y}{b}=0$
$\Rightarrow \frac{b}{y}(1-\frac{x^2}{y^2})-\frac{y}{b}=0$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which is an ellipse.
Option B,
$x=\frac{a}{2}(t+\frac{1}{t})$

$\Rightarrow \frac{2x}{a}=t+\frac{1}{t}$ .....(iii)
$y=\frac{b}{2}(t-\frac{1}{t})$
$\Rightarrow \frac{2y}{b}=t-\frac{1}{t}$ .....(iv)
$\Rightarrow \frac{{4x}^2}{a^2}-\frac{{4y}^2}{b^2}=4$
$\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
which represents a hyperbola
Option C,
$x=e^t+e^{-t}$ ....(v)

$y=e^t-e^{-t}$ (vi)

$\Rightarrow x^2-y^2=4$

which is a hyperbola
Option D,
$x^2=2(\cos t+3)$ .....(vii)

$\Rightarrow x^2=4{\cos }^2\frac{t}{2}+4$
$\Rightarrow \frac{x^2}{4}={\cos }^2\frac{t}{2}+1$
$y^2=2\left({\cos }^2\frac{t}{2}-1\right)$ .....(viii)

$\Rightarrow \frac{y^2}{2}={\cos }^2\frac{t}{2}-1$
So, $\Rightarrow \frac{x^2}{8}-\frac{y^2}{4}=1$

which represents a hyperbola.

Hence, option A.