Question

For the hyperbola $9x^2-16y^2-18x+32y-151=0$

• A. one of the directrix is $x=\frac{21}{5}$
• B. length of latus rectum$=\frac{9}{2}$
• C. foci are $(6,1)$ and $(-4,1)$
• D. eccentricity is $\frac{5}{4}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A one of the directrix is $x=\frac{21}{5}$
B length of latus rectum$=\frac{9}{2}$
C foci are $(6,1)$ and $(-4,1)$
D eccentricity is $\frac{5}{4}$

The given hyperbola can be written as
$\frac{(x-1{)}^2}{16}-\frac{(y-1{)}^2}{9}=1$
or $\frac{X^2}{16}-\frac{Y^2}{9}=1$
(where $X=x-1,Y=y-1$)
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
Directrices are $X=\pm \frac{a}{e}$. Therefore,
$x-1=\pm \frac{16}{5}$
or $x=\frac{21}{5}$ and $x=-\frac{11}{5}$
Length of latus rectum $=\frac{2b^2}{a}=\frac{9}{2}$
The foci are given by
$X=\pm ae,Y=0$
or $(6,1),(-4,1)$