Question

Let f be a strictly decreasing function defined on R such that f(x)>0,∀x∈R. Let $\frac{x^2}{f(a^2+5a+3)}+\frac{y^2}{f(a+15)}=1$ be an ellipse with a major axis along the y-axis. The value of 'a' can lie in the interval(s)

• A. (-∞,-6)
• B. (-6,2)
• C. (2,∞)
• D. (-∞,∞)

Answer 1

Answer: (A), (C)

(-∞,-6)

Answer 2

The given equation represents an ellipse centered at the origin, with its major axis along the y-axis. The standard form equation of such an ellipse is:

x2/b2+y2+a2​=1

where a is the length of the semi-major axis and b is the length of the semi-minor axis.

Comparing the given equation (2+5+3)2+(+15)2=1 f(a 2+5 a +3)x 2+f(a +15)y 2=1 to the standard form, we see that 2+5+3 a 2+5 a +3 corresponds to a 2 and a+15 a corresponds to b 2.

Given that f is a strictly decreasing function and f(x)>0 for all x , we can conclude that a 2+5 a +3 must be greater than a +15 for the major axis to be along the y-axis. This leads to the inequality 2+4−12>0 a 2+4 a −12>0, which factors as (2)(6)>0(a −2)(a +6)>0.

This implies that a must lie in the interval (−∞,−6)∪(2,∞)(−∞,−6)∪(2,∞) for the major axis to be along the y-axis and for the given equation to represent an ellipse. Hence, the intervals (−∞,−6)(−∞,−6) and (2,∞)(2,∞) are the correct intervals for the value of a.

The correct answer is/are option(s):
(A): (-∞,-6)
(C): (2,∞)