Question

The length of the latus rectum of the curve traced by the point $(\frac{3}{2}[t+\frac{1}{t}],\frac{\sqrt{7}}{2}[t-\frac{1}{t}])$, where $t(\ne 0)$ is a parameter, is equal to

• A. $\frac{6}{7}$
• B. $\frac{14}{3}$
• C. $\frac{7}{3}$
• D. $\frac{3}{7}$

Answer 1

The correct option is B $\frac{14}{3}$

Let $x=\frac{3}{2}[t+\frac{1}{t}]$ and
$\Rightarrow \frac{2x}{3}=t+\frac{1}{t}\dots \left(1\right)$
$y=\frac{\sqrt{7}}{2}[t-\frac{1}{t}]$
$\frac{2y}{\sqrt{7}}=t-\frac{1}{t}\dots \left(2\right)$
From $(1{)}^2-(2{)}^2$
$\Rightarrow {\left(\frac{2x}{3}\right)}^2-{\left(\frac{2y}{\sqrt{7}}\right)}^2=4$
$\Rightarrow \frac{x^2}{9}-\frac{y^2}{7}=1$, which is a hyperbola
$\therefore$ Length of latus rectum $=\frac{2b^2}{a}=\frac{14}{3}$