Question ID: #880
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
- (1) 516
- (2) 256
- (3) 496
- (4) 276
Solution:
$$f(t) = -\frac{3}{4} + 2t – t^2 = -\left(t^2 – 2t + \frac{3}{4}\right)$$
To find maximum value, perfect square method or differentiation:
$$f(t) = -\left(t^2 – 2t + 1 – 1 + \frac{3}{4}\right) = -(t-1)^2 + 1 – \frac{3}{4} = -(t-1)^2 + \frac{1}{4}$$
$$f(t)_{max} = \frac{1}{4} \Rightarrow e = \frac{1}{4}$$
Relation between $a, b, e$: $e^2 = 1 – \frac{b^2}{a^2}$
$$\left(\frac{1}{4}\right)^2 = 1 – \frac{b^2}{a^2} \Rightarrow \frac{1}{16} = \frac{a^2 – b^2}{a^2} \Rightarrow a^2 – b^2 = \frac{a^2}{16}$$
$$15a^2 = 16b^2$$
Length of Latus Rectum: $\frac{2b^2}{a} = 30 \Rightarrow b^2 = 15a$
Substitute $b^2$:
$$15a^2 = 16(15a) \Rightarrow a^2 = 16a \Rightarrow a = 16 \quad (\because a \neq 0)$$
$$a^2 = 256$$
$$b^2 = 15(16) = 240$$
$$a^2 + b^2 = 256 + 240 = 496$$
Ans. (3)
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