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

The Correct Option is A, C

The given equation represents an ellipse centered at the origin, with its major axis along the y-axis.

The standard form of the ellipse is:

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$  

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

Now, consider the given equation:

$f(a^2 + 5a + 3) \cdot x^2 + f(a + 15) \cdot y^2 = 1$

To match the standard form, we divide both sides of the equation by 1 (which is already done), and compare:

$\frac{x^2}{\frac{1}{f(a^2 + 5a + 3)}} + \frac{y^2}{\frac{1}{f(a + 15)}} = 1$ 

So, we can see: 
$b^2 = \frac{1}{f(a^2 + 5a + 3)}$, 
$a^2 = \frac{1}{f(a + 15)}$

Now, we are told that $f$ is strictly decreasing and $f(x) > 0$ for all $x$. 
Since the major axis is along the y-axis, it must be that:

$\frac{1}{f(a + 15)} > \frac{1}{f(a^2 + 5a + 3)}$ 

Because $f(x)$ is strictly decreasing, the above inequality holds when: 
$a^2 + 5a + 3 > a + 15$

Solving this: 
$a^2 + 5a + 3 > a + 15$ 
$\Rightarrow a^2 + 4a - 12 > 0$ 
$\Rightarrow (a - 2)(a + 6) > 0$

This inequality holds when: 
$a \in (-\infty, -6) \cup (2, \infty)$

Conclusion: For the given equation to represent an ellipse with the major axis along the y-axis, the value of $a$ must lie in the intervals:

(A): $(-\infty, -6)$
(C): $(2, \infty)$