The eccentricity of the conic given by $x = t^{2} + t + 1,$ $y = t^{2} - t + 1$ is $e,$ where $t$ is a parameter. Then $e^{2}$ is equal to

Open in App

Solution

As $\frac{x - y}{2} = t,$
$\Rightarrow x = {(\frac{x - y}{2})}^{2} + (\frac{x - y}{2}) + 1$
$\Rightarrow {(\frac{x - y}{2})}^{2} = (\frac{x + y}{2}) - 1$
which represents a parabola.
Hence, $e = 1$

Suggest Corrections

Similar questions

x = t^{2} + t + 1

y = t^{2} - t + 1

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

y = \frac{2 a t}{1 + t^{2}}

t

x = \frac{1 - t^{2}}{1 + t^{2}}, y = \frac{2 t}{1 + t^{2}}

t

x + y = t - \frac{1}{t}, x^{2} + y^{2} = t^{2} + \frac{1}{t^{2}},

\frac{d y}{d x}

t

x = a \frac{1 - t^{2}}{1 + t^{2}}, y = b \frac{2 t}{1 + t^{2}}

Join BYJU'S Learning Program