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Question #40973

Difficulty: 8/10 | 🟢 Free

The Question

Consider the family of ellipses $E_t: \frac{x^2}{t} + \frac{y^2}{10-t} = 1$ , where $t \in [2, 8]$ . For each $t$ , let $D_t$ be the diameter of $E_t$ that bisects all chords of slope $1$ . Let $P_t$ be the point on $D_t$ whose abscissa is $\frac{t}{10}$ . As $t$ varies, $P_t$ traces a path on a straight line $\Gamma$ . For each $t$ , let $C_t$ be the chord of $E_t$ that lies on the line $\Gamma$ . Let $L(t)$ be the length of the chord $C_t$ . If $R$ is the range of $L(t)$ and $C_R$ is the coefficient of range of $L(t)$ for $t \in [2, 8]$ , find the value of $R \cdot C_R$ .

Options

A.
$\frac{\sqrt{2}}{15}$
B.
0
C.
$3\sqrt{2}$
D.
$\frac{2\sqrt{2}}{15}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Range and Coefficient of Range|Diameter of Conics|Locus of a Moving Point

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to link the locus of a diameter-dependent point to the optimization of a geometric property (chord length).

The primary trap lies in the range calculation of the chord length function $L(t) \propto \sqrt{10t-t^2}$ over the closed interval $[2, 8]$ .

Students often erroneously assume that the extrema of a function on a closed interval occur strictly at the boundaries.

By evaluating $L(t)$ only at $t=2$ and $t=8$ , they will find identical values, leading to a false Range of 0.

The correct approach requires identifying the interior global maximum at $t=5$ using derivatives, thereby correctly distinguishing between the parameter's domain and the geometric function's codomain.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Statistics - Class 11
Concept	Range and Coefficient of Range
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free