Question

The equation of the Latus Rectum of the hyperbola $\frac{{(x-4)}^2}{16}-\frac{{(y-3)}^2}{20}=1$ are____.

• A. $x=1\pm 5.$
• B. $y=2\pm 6.$
• C. $x=4\pm 6.$
• D. $y=3\pm 5.$

Answer 1

The correct option is C

$x=4\pm 6.$

Find the equation of the latus rectum of the given hyperbola.

We are given the hyperbola is $\frac{{(x-4)}^2}{16}-\frac{{(y-3)}^2}{20}=1$

Where, $a=4,andb=2\sqrt{5}$

$$\begin{gathered} e=\frac{\sqrt{a^2+b^2}}{a} \\ =\frac{\sqrt{16+20}}{4} \end{gathered}$$

$x-x_1=\pm ac$ [The equation of the latus rectum of the hyperbola]

We know that $x_1=4$ [ $x_1=$ The coordinate of the x-axis]

$$\begin{gathered} x=x_1\pm ac \\ =4\pm 4\left(\frac{3}{2}\right) \\ x=4\pm 6 \end{gathered}$$

Hence, the equation of the latus rectum of the given hyperbola is $x=4\pm 6.$