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 : [2025]

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Solution

Q.
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 : [2025]

1 $\frac{25}{6}$

2 $\frac{24}{5}$

3 $\frac{144}{5}$

4 $\frac{288}{5}$

Ans.
(4)

Distance between foci $\sqrt{{(1 - 1)}^{2} + {(14 + 12)}^{2}}$

$\Rightarrow 2 b e = 26 \Rightarrow b e = 13$

Mid-point of foci $= (\frac{1 + 1}{2}, \frac{14 - 12}{2}) = (1, 1)$

Equation of hyperbola is $\frac{{(x - 1)}^{2}}{a^{2}} + \frac{{(y - 1)}^{2}}{b^{2}} = 1$

Hyperbola passes through (1, 6)

$∴ \frac{{(6 - 1)}^{2}}{b^{2}} = 1 \Rightarrow b^{2} = 25$

Now, $a^{2} = b^{2} (e^{2} - 1) = b^{2} e^{2} - b^{2} = 169 - 25 = 144$

$∴ a^{2} = 144 and b^{2} = 25$

$∴$ Length of latus rectum $= \frac{2 a^{2}}{b} = \frac{2 \times 144}{5} = \frac{288}{5}$

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