Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t

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Question

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$, then $(a^2 + b^2)$ is equal to

276

516

256

496

The correct answer is

516

The problem asks for the value of $a^2 + b^2$ for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$), given the length of its latus rectum and its eccentricity.

Eccentricity Calculation

The eccentricity ($e$) of the ellipse is given as the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$. This is a quadratic function representing a parabola opening downwards. The maximum value occurs at the vertex.

The t-coordinate of the vertex is calculated using $t = -\frac{B}{2A}$, where $A=-1$ and $B=2$ are the coefficients of $t^2$ and $t$ respectively. $t = -\frac{2}{2 \times (-1)} = 1$.
Substitute $t=1$ back into the function to find the maximum value: $f(1) = -\frac{3}{4} + 2(1) - (1)^2 = -\frac{3}{4} + 2 - 1 = 1 - \frac{3}{4} = \frac{1}{4}$.
Therefore, the eccentricity of the ellipse is $e = \frac{1}{4}$.

Latus Rectum Information

The length of the latus rectum ($LR$) of the ellipse is given as 30.

The formula for the latus rectum of an ellipse with $a > b$ is $LR = \frac{2b^2}{a}$.
Setting the given value: $\frac{2b^2}{a} = 30$.
Simplifying gives $\frac{b^2}{a} = 15$.

Solving for Ellipse Parameters

We use the relationship between eccentricity ($e$), semi-major axis ($a$), and semi-minor axis ($b$) for an ellipse where $a > b$: $e^2 = 1 - \frac{b^2}{a^2}$.

Substitute the value of $e = \frac{1}{4}$: $(\frac{1}{4})^2 = 1 - \frac{b^2}{a^2}$ $\frac{1}{16} = 1 - \frac{b^2}{a^2}$.
Rearrange to find $\frac{b^2}{a^2}$: $\frac{b^2}{a^2} = 1 - \frac{1}{16} = \frac{15}{16}$.
We now have two equations: 1) $\frac{b^2}{a} = 15 \implies b^2 = 15a$ 2) $\frac{b^2}{a^2} = \frac{15}{16} \implies b^2 = \frac{15}{16}a^2$
Substitute $b^2$ from the first equation into the second: $15a = \frac{15}{16}a^2$.
Since $a$ must be positive (length of semi-major axis), we can divide by $15a$: $1 = \frac{1}{16}a \implies a = 16$.
Now, find $b^2$ using $b^2 = 15a$: $b^2 = 15 \times 16 = 240$.

Final Calculation

The question asks for the value of $a^2 + b^2$.

Calculate $a^2$: $a^2 = 16^2 = 256$.
Calculate $a^2 + b^2$: $a^2 + b^2 = 256 + 240 = 496$.

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