What condition gives major axis along y for x²/f(a²+7a+3)+y²/f(3a+15)=1?

f(3a+15)>f(a²+7a+3). Since f is strictly decreasing, this gives 3a+15 0, so a 2.

What is α²+β² where valid set is ℝ−[α,β]?

α=−6, β=2, α²+β²=36+4=40.

Ellipse x²/f(a²+7a+3)+y²/f(3a+15)=1 major axis along y-axis, f strictly decreasing positive — set of all a is ℝ−[α,β], find α²+β²

JEERANKERS

MathEllipse

QMCQConic Sections

Let $\dfrac{x^2}{f(a^2+7a+3)}+\dfrac{y^2}{f(3a+15)}=1$ represent an ellipse with major axis along the $y$-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) $24$

✅ Correct Answer

B) 40

✓

Solution

Conditions for the given conic to be an ellipse with major axis along y

For $\dfrac{x^2}{p}+\dfrac{y^2}{q}=1$ to be an ellipse with major axis along $y$-axis:

1. $p>0$ and $q>0$ — satisfied since $f$ is positive everywhere ✓
2. $q>p$ — the $y$-denominator must exceed the $x$-denominator
3. $p\neq q$ — not a circle

Condition 2: $f(3a+15)>f(a^2+7a+3)$.

Apply the strictly decreasing property of f

Since $f$ is strictly decreasing: $f(u)>f(v)\Leftrightarrow u

$f(3a+15)>f(a^2+7a+3)$

$\Leftrightarrow\quad 3a+15 < a^2+7a+3$

$\Leftrightarrow\quad 0 < a^2+4a-12$

$\Leftrightarrow\quad (a+6)(a-2)>0$

$\Leftrightarrow\quad a<-6\quad\text{or}\quad a>2$

Identify α, β and compute α²+β²

The set of valid $a$ values is $(-\infty,-6)\cup(2,\infty)=\mathbb{R}-[-6,2]$.

So $[\alpha,\beta]=[-6,2]$, giving $\alpha=-6$ and $\beta=2$.

$$\alpha^2+\beta^2=(-6)^2+2^2=36+4=\boxed{40}$$

📘 Ellipse Orientation Condition with Decreasing Function

For $x^2/p+y^2/q=1$ with major axis along $y$: need $q>p>0$. When $p=f(u)$ and $q=f(v)$ with $f$ strictly decreasing, $f(v)>f(u)\Leftrightarrow v

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Important Concepts

Ellipse with Major Axis along yFor $x^2/p+y^2/q=1$: major axis is along $y$-axis iff $q>p>0$. The $y$-denominator must be strictly larger than the $x$-denominator.

Using Strictly Decreasing f$f$ strictly decreasing means $f(u)>f(v)\Leftrightarrow uf(a^2+7a+3)\Leftrightarrow 3a+150$.

Solving a²+4a−12>0Factor: $(a+6)(a-2)>0$. Roots: $a=-6$ and $a=2$. The quadratic is positive outside the roots: $a<-6$ or $a>2$. The excluded interval is $[-6,2]$.

Final Calculation$\alpha=-6$, $\beta=2$. $\alpha^2+\beta^2=36+4=40$. Answer B.

FAQs

Why does major axis along y require q>p?

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The major axis lies along the direction with the larger semi-axis. Semi-axis along $y$ is $\sqrt{q}$ and along $x$ is $\sqrt{p}$. For $y$-major axis: $\sqrt{q}>\sqrt{p}$, i.e., $q>p$.

How does f being decreasing flip the inequality?

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If $f$ is decreasing and $f(A)>f(B)$, then $A$ must be smaller than $B$ (you go further left to get a larger value). So $f(3a+15)>f(a^2+7a+3)\Rightarrow3a+15

What are the roots of a²+4a−12=0?

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$a=\frac{-4\pm\sqrt{16+48}}{2}=\frac{-4\pm8}{2}$. So $a=2$ or $a=-6$.

Why is the excluded set [−6,2] and not (−6,2)?

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At $a=-6$ or $a=2$: $(a+6)(a-2)=0$, meaning $p=q$, giving a circle (not an ellipse). So the endpoints are also excluded.

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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