Question

Let \(\dfrac32\) be the eccentricity of the conjugate hyperbola of a hyperbola \(H\) and one of the foci of \(H\) lie on the straight line \[ x+y-3=0. \] If the transverse and conjugate axes of \(H\) are along \(X\)-axis and \(Y\)-axis respectively, then the length of the latus rectum of \(H\) is

• A. \(\dfrac{4}{\sqrt5}\)
• B. \(\dfrac{8}{\sqrt5}\)
• C. \(\dfrac{10}{3}\)
• D. \(\dfrac53\)

Hint:

For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] \[ \boxed{c^2=a^2+b^2} \] and \[ \boxed{\text{Length of latus rectum}=\frac{2b^2}{a}.} \]

Answer 1

The Correct Option is B

Step 1: Write the equation of the hyperbola. Since the transverse axis is along the \(X\)-axis, \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1. \] Its conjugate hyperbola is \[ \frac{y^2}{b^2}-\frac{x^2}{a^2}=1. \]

Step 2: Use the eccentricity of the conjugate hyperbola. The eccentricity of the conjugate hyperbola is \[ e'=\sqrt{1+\frac{a^2}{b^2}}. \] Given, \[ e'=\frac32. \] Hence, \[ 1+\frac{a^2}{b^2}=\frac94, \] \[ \frac{a^2}{b^2}=\frac54, \] or \[ a^2:b^2=5:4. \]

Step 3: Use the given focus. For the hyperbola, \[ c^2=a^2+b^2. \] Using the ratio, \[ c^2=5k+4k=9k, \] so \[ a=\sqrt{5k}, \qquad b=2\sqrt{k}, \qquad c=3\sqrt{k}. \] One focus is \[ (3\sqrt{k},0). \] Since it lies on \[ x+y-3=0, \] we get \[ 3\sqrt{k}=3, \] so \[ k=1. \] Therefore, \[ a=\sqrt5, \qquad b=2. \]

Step 4: Find the length of the latus rectum. For the hyperbola, \[ \boxed{\text{Length of latus rectum}=\frac{2b^2}{a}.} \] Hence, \[ \frac{2(2)^2}{\sqrt5} = \frac8{\sqrt5}. \] Therefore, \[ \boxed{\frac8{\sqrt5}}. \] Hence, the correct option is \(\boxed{(B)}\).