Let the eccentricity of an ellipse x²/a² + y²/b² = 1, a b, be 1/4. If this ellipse passes through

Q: Wait, why do we use $e^2 = 1 - \frac{b^2}{a^2}$? Can't we just use $b^2 = a^2(1-e^2)$?

Absolutely! They are mathematically identical. The form $e^2 = 1 - \frac{b^2}{a^2}$ is just a rearrangement. Use whichever feels more intuitive to you. The key is to remember that for a horizontal ellipse ($a > b$), the ratio $\frac{b^2}{a^2}$ must be less than 1, which is exactly what $1 - e^2$ provides.

Q: Why did we assume $a > b$ at the start?

The problem explicitly states $a > b$. This is a crucial piece of information because it defines the orientation of the ellipse. If $b > a$, the ellipse would be vertical, and the eccentricity formula would change to $e^2 = 1 - \frac{a^2}{b^2}$. Always check the orientation before you start calculating!

Q: I got stuck squaring the point $(-4\sqrt{\frac{2}{5}}, 3)$. Is there a trick?

It looks intimidating, but take it slow. When you square $-4\sqrt{\frac{2}{5}}$, remember to square the $-4$ (which gives $16$) and the square root separately (which gives $\frac{2}{5}$). Then multiply them: $16 \times \frac{2}{5} = \frac{32}{5}$. Don't rush the arithmetic; precision is your best friend in JEE Advanced.

Question

Let the eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$, be $\frac{1}{4}$. If this ellipse passes through the point $(-4\sqrt{\frac{2}{5}}, 3)$, then $a^2 + b^2$ is equal to :

Options:

(A) 29

(B) 31

(C) 32

(D) 34

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are exploring the geometry of an ellipse.

An ellipse is defined by its own internal tension—the balance between its major and minor axes. When we look at the equation:

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

We are looking at the blueprint of this balance. The condition $a > b$ tells us that our ellipse is stretched along the horizontal axis, a graceful elongation that defines its character.

### The Secret Language of Eccentricity

We are given the eccentricity $e = \frac{1}{4}$. Think of eccentricity as the 'DNA' of the ellipse, telling us exactly how far the ellipse has drifted from the perfect symmetry of a circle.

To connect this geometric property to our algebraic variables, we invoke the fundamental relationship:

$$e^2 = 1 - \frac{b^2}{a^2}$$

By substituting $e = \frac{1}{4}$, we obtain:

$$\left(\frac{1}{4}\right)^2 = 1 - \frac{b^2}{a^2} \Rightarrow \frac{1}{16} = 1 - \frac{b^2}{a^2}$$

With a quick algebraic shuffle, we find that $\frac{b^2}{a^2} = \frac{15}{16}$, or more usefully:

$$b^2 = \frac{15}{16}a^2$$

We have now reduced our two unknowns to a single variable, $a^2$. This is the moment where the problem begins to yield.

### The Moment of Truth: Testing the Point

We are given a point $P\left(-4\sqrt{\frac{2}{5}}, 3\right)$ that lies on the boundary of our ellipse. In the world of coordinate geometry, this is a command: the coordinates must satisfy the equation.

When we substitute these values into the ellipse equation, we get:

$$\frac{\left(-4\sqrt{\frac{2}{5}}\right)^2}{a^2} + \frac{3^2}{b^2} = 1$$

When you square $-4\sqrt{\frac{2}{5}}$, you square the $-4$ to get $16$ and the $\sqrt{\frac{2}{5}}$ to get $\frac{2}{5}$. Multiplying them gives us $\frac{32}{5}$.

So, our equation transforms into:

$$\frac{32}{5a^2} + \frac{9}{b^2} = 1$$

### The Final Convergence

Now, we substitute our expression $b^2 = \frac{15}{16}a^2$ into the equation:

$$\frac{32}{5a^2} + \frac{9}{\frac{15}{16}a^2} = 1$$

The term $\frac{9}{\frac{15}{16}a^2}$ simplifies by moving the $16$ to the numerator:

$$\frac{9 \cdot 16}{15a^2} = \frac{3 \cdot 16}{5a^2} = \frac{48}{5a^2}$$

Now, look at the equation:

$$\frac{32}{5a^2} + \frac{48}{5a^2} = 1$$

Since the denominators are identical, we add the numerators:

$$\frac{80}{5a^2} = 1 \Rightarrow \frac{16}{a^2} = 1 \Rightarrow a^2 = 16$$

With $a^2 = 16$ in hand, finding $b^2$ is trivial:

$$b^2 = \frac{15}{16} \cdot 16 = 15$$

The final step is to calculate $a^2 + b^2$:

$$a^2 + b^2 = 16 + 15 = \mathbf{31}$$

We have navigated the geometry, mastered the algebra, and arrived at the solution. Remember, in JEE Advanced, the math is never just about the numbers; it is about the structure.

Animated Solution for Mathematics - Conic Sections: Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ , be $\frac{1}{4}$ . If this ellipse passes through the point $(- 4 \frac{2}{5}, 3)$ , then $a^{2} + b^{2}$ is equal to :

Select Answer:

Visualized Solution

The Standard Ellipse

Eccentricity Formula

Substitute Eccentricity

Express $b^{2}$ in terms of $a^{2}$

The Point on the Ellipse

Substitute the Point

Simplify the Numerators

Substitute $b^{2}$

Simplify the Fraction

Solve for $a^{2}$

Calculate $b^{2}$

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Secret Language of Eccentricity

The Moment of Truth

The Final Convergence

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