If $\frac{x^{2}}{f (4 a)} + \frac{y^{2}}{f (a^{2} - 5)} = 1$ represents an ellipse with major axis as y-axis and f is a decreasing function, then

$a \in (- \infty, 1)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$a \in (5, \infty)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$a \in (1, 4)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$a \in (- 1, 5)$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$a \in (- 1, 5)$

Since y-axis is major axis
$\Rightarrow f (4 a) < f (a^{2} - 5) \Rightarrow 4 a > a^{2} - 5 (∵ f is decreasing) \Rightarrow a^{2} - 4 a - 5 < 0 \Rightarrow a \in (- 1, 5)$

Suggest Corrections

Similar questions

If $\frac{x^{2}}{f (4 a)} + \frac{y^{2}}{f (a^{2} - 5)} = 1$ represents an ellipse with major axis as y-axis and f is a decreasing function, then

\frac{x^{2}}{f (4 a)} = \frac{y^{2}}{f (a^{2} - 5)}

f

\frac{x^{2}}{a^{2} - 5} + \frac{y^{2}}{4 a} = 1

y -

\frac{x^{2}}{3 - α} + \frac{y^{2}}{2} = 1

x -

α

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

Join BYJU'S Learning Program