If the foci of a hyperbola are same as that of the ellipse x²/9 + y²/25 = 1 and the eccentricity of

Q: Wait, why do we assume the hyperbola is vertical? Couldn't it be horizontal?

That is a fantastic question! Remember, the problem explicitly states that the hyperbola shares the *same* foci as the ellipse. Since the ellipse's foci lie on the $y$-axis at $(0, \pm 4)$, the hyperbola's foci must also lie on the $y$-axis. By definition, the transverse axis of a hyperbola contains its foci. Therefore, the hyperbola must be vertical (a conjugate hyperbola).

Q: How do I know whether to use $ey - B$ or $ey + B$ for the smaller distance?

Think about the geometry. The foci are at $(0, 4)$ and $(0, -4)$. For a point $P$ with a positive $y$-coordinate, it is closer to the upper focus $(0, 4)$. The distance formula for the upper focus is $|ey - B|$. Since $ey$ is the distance from the origin to the projection of the point on the directrix-related line, subtracting $B$ gives the distance to the closer focus. If you are ever unsure, calculate both and pick the smaller one!

Q: Why can't we just use the distance formula $\sqrt{(x-x_f)^2 + (y-y_f)^2}$?

You absolutely *can*! However, that path is algebraically treacherous. You would be squaring roots and dealing with complex radicals. Using the property that the focal distance of a point $(x_1, y_1)$ on a hyperbola is $|ey_1 \pm B|$ is a powerful tool that saves you time and reduces the chance of calculation errors. In JEE Advanced, recognizing these properties is what separates the good from the great.

Question

If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $(\sqrt{2}, \frac{14}{3}\sqrt{\frac{2}{5}})$ on the hyperbola, is equal to

Options:

(A) $7\sqrt{\frac{2}{5}} - \frac{8}{3}$

(B) $14\sqrt{\frac{2}{5}} - \frac{4}{3}$

(C) $14\sqrt{\frac{2}{5}} - \frac{16}{3}$

(D) $7\sqrt{\frac{2}{5}} + \frac{8}{3}$

Sigma Percentile · Accepted Answer

### The Dance of the Conics: A Journey Through Symmetry

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are embarking on a journey of symmetry. We are going to bridge the gap between the closed, comforting world of the ellipse and the infinite, branching paths of the hyperbola.

This problem is a beautiful reminder that in mathematics, everything is connected. If you understand the soul of a conic section, you don't need to memorize formulas—you simply need to visualize the geometry.

### Phase 1: Decoding the Ellipse

Let us begin with the ellipse:

$$\frac{x^2}{9} + \frac{y^2}{25} = 1$$

When you look at this equation, you see a shape stretched along the $y$-axis. Because the denominator under $y^2$ is $25$, which is greater than the $9$ under $x^2$, our major axis is vertical.

To find the eccentricity $e_E$, we use the standard relation for a vertical ellipse:

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

Substituting our values, we get:

$$e_E = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

With this, we can locate the foci. For a vertical ellipse, the foci are at $(0, \pm be_E)$. Since $b = 5$ and $e_E = \frac{4}{5}$, the foci are at $(0, \pm 4)$. This is our anchor point, and the hyperbola we are about to construct will share these exact coordinates.

### Phase 2: The Hyperbola's Identity

The problem states that the hyperbola's eccentricity $e_H$ is $\frac{15}{8}$ times that of the ellipse. This is a direct calculation:

$$e_H = \frac{15}{8} 	imes \frac{4}{5}$$

The $5$ goes into $15$ three times, and $4$ goes into $8$ two times. We are left with $e_H = \frac{3}{2}$.

Since the hyperbola shares the foci $(0, \pm 4)$, it must also be vertical. Its equation takes the form:

$$\frac{y^2}{B^2} - \frac{x^2}{A^2} = 1$$

In this configuration, the distance from the center to the focus is given by $Be_H$. We know this distance must be $4$, so $B \cdot \frac{3}{2} = 4$, which gives us $B = \frac{8}{3}$. We have now defined the hyperbola's skeleton.

### Phase 3: The Focal Distance Climax

We are given a point $P(\sqrt{2}, \frac{14}{3}\sqrt{\frac{2}{5}})$ on this hyperbola. We need the smaller focal distance.

For a vertical hyperbola, the focal distances of a point $(x_1, y_1)$ are simply $|ey_1 \pm B|$. Since our point has a positive $y$-coordinate, it lies on the upper branch of the hyperbola and is closer to the upper focus $(0, 4)$. Therefore, the smaller distance is $e_H y_1 - B$.

Let us perform the final substitution with care:

$$d = \left(\frac{3}{2}ight) \left(\frac{14}{3}\sqrt{\frac{2}{5}}ight) - \frac{8}{3}$$

The $3$ in the numerator and the $3$ in the denominator cancel out perfectly. We are left with $\frac{14}{2} \sqrt{\frac{2}{5}}$, which simplifies to $7\sqrt{\frac{2}{5}}$.

Our final result is:

$$\mathbf{7\sqrt{\frac{2}{5}} - \frac{8}{3}}$$

### Reflection

See how the complexity melted away? We didn't need to brute-force the distance formula. We used the geometric properties of the conic sections to guide us.

Always remember: in JEE Advanced, the path of least resistance is usually the one paved with conceptual understanding. You have successfully navigated from an ellipse to a hyperbola and found the distance to its focus. Keep this clarity of thought, and no problem will ever be too daunting.

Select Answer:

Visualized Solution

Analyze the Ellipse Equation

Calculate Ellipse Eccentricity $e_{E}$

Locate the Foci of the Ellipse

Determine Hyperbola Eccentricity $e_{H}$

Identify Hyperbola Type and Parameter $B$

Define Focal Distance for Hyperbola

Raw Setup (Substitution)

Atomic Compute (Execution)

Final Conclusion

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

The Dance of the Conics

Phase 1

Phase 2

Phase 3

Reflection

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