QJEEM24Madb69: JEE Main 2024 Mathematics - Class 11 JEEM24Madb69 Solution: Let be an ellipse, whose eccentricity is and the length of the…

Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with a > b , the eccentricity $e$ is given by the formula:

$e = \sqrt{1 - \left(\frac{b}{a}\right)^2}$

It is provided that the eccentricity $e$ is $\frac{1}{\sqrt{2}}$ (given), so we can equate the two expressions for eccentricity:

$\frac{1}{\sqrt{2}} = \sqrt{1 - \left(\frac{b}{a}\right)^2}$

Squaring both sides to eliminate the square root gives:

$\frac{1}{2} = 1 - \left(\frac{b}{a}\right)^2$

$\left(\frac{b}{a}\right)^2 = 1 - \frac{1}{2}$

$\left(\frac{b}{a}\right)^2 = \frac{1}{2}$

Taking the square root on both sides:

$\frac{b}{a} = \frac{1}{\sqrt{2}}$

$a = b\sqrt{2}$

Now, for the ellipse, the length of the latus rectum is given by the formula:

$\text{Length of Latus Rectum (L)} = \frac{2b^2}{a}$

It's provided that the length of the latus rectum $L$ is $\sqrt{14}$ , so substitute the known values to find $b$ :

$\sqrt{14} = \frac{2b^2}{b\sqrt{2}} = \frac{2b}{\sqrt{2}}$

$b\sqrt{2} = \sqrt{14}$

$b^2 = \frac{14}{2}$

$b^2 = 7$

And since $a = b\sqrt{2}$ , we can find $a^2$ :

$a^2 = (b\sqrt{2})^2$

$a^2 = 7 \cdot 2$

$a^2 = 14$

Now we have an ellipse with $a^2 = 14$ and $b^2 = 7$ . The equation of a hyperbola similar to the given ellipse but with the terms subtracted is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

For the hyperbola, the square of the eccentricity $e'$ is given by:

$(e')^2 = 1 + \frac{b^2}{a^2}$

Substitute the values we've found for $a^2$ and $b^2$ into the formula for the square of the hyperbola's eccentricity:

$(e')^2 = 1 + \frac{b^2}{a^2}$

$(e')^2 = 1 + \frac{7}{14}$

$(e')^2 = 1 + \frac{1}{2}$

$(e')^2 = \frac{3}{2}$

Therefore, the square of the eccentricity of the hyperbola is $\frac{3}{2}$ , which corresponds to option C.