Let e1 and e2 be the eccentricities of the ellipse x²/b² + y²/25 = 1 and the hyperbola x²/16 - y²/b²

Q: Wait, why is the ellipse vertical? The equation looks like a standard horizontal one.

Great catch! In the equation $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$, the denominator of $y^2$ is $25$. Since we are given $b b^2$, the major axis lies along the $y$-axis, making it a vertical ellipse.

Q: Why is the hyperbola eccentricity formula $1 + \frac{b^2}{16}$? I thought it was $1 - \dots$?

That is a very common confusion! Remember, the ellipse is a 'bounded' shape, so its eccentricity must be less than $1$, hence the subtraction ($1 - \frac{b^2}{a^2}$). The hyperbola, however, is 'unbounded' and spreads out, so its eccentricity must be greater than $1$. Thus, we use the addition formula ($1 + \frac{b^2}{a^2}$). Keep this conceptual difference in mind!

Q: How do we know the foci are the vertices of the new ellipse?

The problem states the new ellipse passes through the four foci. Since the foci of the first ellipse are at $(0, \pm 4)$ and the foci of the hyperbola are at $(\pm 5, 0)$, these points lie directly on the coordinate axes. For an ellipse centered at the origin with axes along the coordinate axes, points on the axes are, by definition, the vertices.

Question

Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{25} = 1$ and the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{b^{2}} = 1$, respectively. If $b < 5$ and $e_{1}e_{2} = 1$, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

Options:

(A) $\frac{4}{5}$

(B) $\frac{3}{5}$

(C) $\frac{\sqrt{7}}{4}$

(D) $\frac{\sqrt{3}}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Orientation

We begin with the ellipse $E_1$ defined by the equation:

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

Given the condition $b < 5$, we observe that $25 > b^2$. In the geometry of ellipses, the major axis is determined by the larger denominator.

Since the larger denominator is associated with the $y$-term, the major axis lies along the $y$-axis. Recognizing this **vertical orientation** is critical to avoiding sign errors in subsequent calculations.

### The Eccentricity Dance

Next, we define the eccentricities for both conic sections. For the vertical ellipse $E_1$, the eccentricity $e_1$ satisfies:

$$e_1^2 = 1 - \frac{b^2}{25}$$

For the hyperbola $H_1$ defined by $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$, the eccentricity $e_2$ satisfies:

$$e_2^2 = 1 + \frac{b^2}{16}$$

We are given the condition $e_1 e_2 = 1$. Squaring both sides yields $e_1^2 e_2^2 = 1$. Substituting our expressions, we obtain:

$$\left(1 - \frac{b^2}{25}ight) \left(1 + \frac{b^2}{16}ight) = 1$$

Expanding the product, we get:

$$1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1$$

The constant terms cancel out, simplifying the equation to:

$$\frac{b^2}{16} - \frac{b^2}{25} = \frac{b^4}{400}$$

Multiplying by $400$ to clear the denominators, we find $25b^2 - 16b^2 = b^4$, which simplifies to $9b^2 = b^4$. Since $b 
eq 0$, we conclude that **$b^2 = 9$**.

### Geometric Synthesis

With $b^2 = 9$, we determine the foci of the original curves. For the ellipse, $e_1^2 = 1 - \frac{9}{25} = \frac{16}{25}$, implying $e_1 = \frac{4}{5}$. The foci are located at $(0, \pm 5 \cdot \frac{4}{5}) = (0, \pm 4)$.

For the hyperbola, $e_2^2 = 1 + \frac{9}{16} = \frac{25}{16}$, implying $e_2 = \frac{5}{4}$. The foci are located at $(\pm 4 \cdot \frac{5}{4}, 0) = (\pm 5, 0)$.

These four points—$(0, 4), (0, -4), (5, 0),$ and $(-5, 0)$—serve as the vertices for a new ellipse. The equation for this ellipse is:

$$\frac{x^2}{25} + \frac{y^2}{16} = 1$$

### Final Calculation

Finally, we calculate the eccentricity $e$ of this new ellipse. With $a^2 = 25$ and $b^2 = 16$, we apply the standard formula:

$$e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

The eccentricity of the resulting ellipse is **$\frac{3}{5}$**.

Select Answer:

Visualized Solution

Equations of $E_{1}$ and $H_{1}$

Eccentricity of $E_{1}$

Eccentricity of $H_{1}$

Applying $e_{1} e_{2} = 1$

Expanding the Equation

Solving for $b^{2}$

Foci of Ellipse $E_{1}$

Foci of Hyperbola $H_{1}$

The New Ellipse Equation

Final Eccentricity Calculation

Summary and Key Takeaway

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Orientation

The Eccentricity Dance

Geometric Synthesis

Final Calculation

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