Question

Let the foci of a hyperbola be \( (1, 14) \) and \( (1, -12) \). If it passes through the point \( (1, 6) \), then the length of its latus-rectum is:

• A. \( \frac{25}{6} \)
• B. \( \frac{24}{5} \)
• C. \( \frac{288}{5} \)
• D. \( \frac{144}{5} \)

Hint:

The length of the latus-rectum of a hyperbola is determined by the formula \( \frac{2a^2}{b} \), where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

Answer 1

The Correct Option is C

The foci of the hyperbola are at \( (1, 14) \) and \( (1, -12) \). The distance between the foci is: \[ be = 13, \quad b = 5 \] From the formula \( a^2 = b^2 (e^2 - 1) \), we get: \[ a^2 = b^2 e^2 - b^2 \] \[ a^2 = 169 - 25 = 144 \] The length of the latus-rectum is given by: \[ \ell (LR) = \frac{2a^2}{b} = \frac{2 \times 144}{5} = \frac{288}{5} \]

Answer 2

The foci lie on the vertical line \(x=1\), so the transverse axis is vertical. Put the centre at the midpoint of the foci: \[ \text{centre }=(1,\tfrac{14+(-12)}{2})=(1,1). \]

Distance between foci \(=26\), so \(2c=26\Rightarrow c=13\). For a vertical transverse hyperbola with centre \((1,1)\) we use the form \[ \frac{(y-1)^2}{a^2}-\frac{(x-1)^2}{b^2}=1, \] with \(c^2=a^2+b^2\).

The point \((1,6)\) lies on the hyperbola. Substitute \(x=1\), \(y=6\): \[ \frac{(6-1)^2}{a^2}-0=1 \quad\Rightarrow\quad \frac{25}{a^2}=1 \Rightarrow a^2=25\ (a=5). \]

Now \(b^2=c^2-a^2=13^2-5^2=169-25=144\).

For the hyperbola \(\dfrac{(y-1)^2}{a^2}-\dfrac{(x-1)^2}{b^2}=1\), the length of each latus-rectum is \[ \text{latus-rectum}=\frac{2b^2}{a}. \] Substitute \(b^2=144,\ a=5\): \[ \text{length}=\frac{2\cdot144}{5}=\frac{288}{5}. \]

Answer

\(\dfrac{288}{5}\). (Option 3)