Question

If the length of latus rectum of a hyperbola whose eccentricity is $\frac{3}{\sqrt{5}}$, centre $(4,3)$ and axis is parallel to coordinate axis is $8$ units, then the equation(s) of the hyperbola is/are

• A. $\frac{(y-3{)}^2}{25}-\frac{(x-4{)}^2}{20}=1$
• B. $\frac{(x-4{)}^2}{25}-\frac{(y-3{)}^2}{20}=1$
• C. $\frac{(x-4{)}^2}{25}-\frac{(y-3{)}^2}{15}=1$
• D. $\frac{(2x+y-11{)}^2}{25}-\frac{(x-2y+2{)}^2}{15}=1$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{(y-3{)}^2}{25}-\frac{(x-4{)}^2}{20}=1$
B $\frac{(x-4{)}^2}{25}-\frac{(y-3{)}^2}{20}=1$

Here $\frac{2b^2}{a}=8$ and
$\frac{3}{\sqrt{5}}=\sqrt{1+\frac{b^2}{a^2}}$
$\Rightarrow \frac{4}{5}=\frac{b^2}{a^2}$
$\Rightarrow a=5,b=2\sqrt{5}$

Thus the equations of the given hyperbola is, if transverse axis parallel to $x$-axis and conjugate axis parallel to $y$-axis:
$\frac{(x-4{)}^2}{25}-\frac{(y-3{)}^2}{20}=1$

The equation of hyperbola if transverse axis is parallel to $y$-axis and conjugate axis parallel to $x$-axis is:
$\frac{(y-3{)}^2}{25}-\frac{(x-4{)}^2}{20}=1$