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

Answer 1

The correct option is B $x=4\pm 6$

We are given the hyperbola

$\frac{x-4{)}^2}{16}-\frac{(y-3{)}^2}{20}=1$ having the Focus at $(\pm ae,3)$

Where $a=4\&b=2\sqrt{5}$

$e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{16+20}}{4}=\frac{3}{2}$

Sp, the lats rectum is the line through the focus and parallel to the directer

$x-x_1=\pm ae$

here $x_1$ is nothing but $4$.

$x=x_1\pm ae=4\pm 4.\left(\frac{3}{2}\right)=4\pm 6$

$x=4\pm 6$