Let x^2a^2+y^2b^2=1,ab be an ellipse, whose eccentricity is 1√2 and the length of the lat…

Question

Let $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ is :

Accepted Answer

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:

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

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

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

Squaring both sides to eliminate the square root gives:

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

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

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

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:

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

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

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

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

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

${b}^{2}=7$

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.

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