JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\dfrac{x^2}{f(a^2+7a+3)} + \dfrac{y^2}{f(3a+15)} = 1\) represent an ellipse with major axis along \(y\)-axis, where \(f\) is a strictly decreasing positive function on \(\mathbb{R}\). If the set of all possible values of \(a\) is \(\mathbb{R} - [\alpha, \beta]\), then \(\alpha^2+\beta^2\) is equal to:
- A \(28\)
- B \(40\)
- C \(61\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(40\)
Step-by-step Solution
Detailed explanation
For the given equation to represent an ellipse with its major axis along the \(y\)-axis, the denominator of \(y^2\) must be strictly greater than the denominator of \(x^2\). \(f(3a+15) > f(a^2+7a+3)\) Since \(f\) is given as a strictly decreasing positive function on…
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