Let a hyperbola passes through the focus of the ellipse x²/25 + y²/16 = 1. The transverse and conjug

Q: Wait, why are the ellipse's foci the hyperbola's vertices? The problem says it 'passes through' the focus.

That is a great observation! The problem states the hyperbola passes through the foci of the ellipse $(\pm 3, 0)$. Because the hyperbola's transverse axis coincides with the ellipse's major axis (the x-axis), and the hyperbola is centered at the origin, these points $(\pm 3, 0)$ must lie on the hyperbola. Since they lie on the transverse axis, they are by definition the vertices of the hyperbola.

Q: Why is the product of eccentricities $e_e \cdot e_h = 1$ so important?

It is the key that unlocks the hyperbola's geometry! We know the ellipse's eccentricity $e_e$ is $3/5$. The condition $e_e \cdot e_h = 1$ forces the hyperbola's eccentricity $e_h$ to be the reciprocal, $5/3$. This is a beautiful property that links the two curves, allowing us to determine the hyperbola's shape entirely from the ellipse's parameters.

Q: Can we assume the hyperbola is horizontal?

Yes, absolutely. The problem explicitly states that the transverse and conjugate axes of the hyperbola coincide with the major and minor axes of the ellipse. Since the ellipse is horizontal (its major axis is along the x-axis), the hyperbola must also be horizontal.

Question

Let a hyperbola passes through the focus of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then

Options:

(A) the equation of hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$

(B) the equation of hyperbola is $\frac{x^2}{9} - \frac{y^2}{25} = 1$

(C) focus of hyperbola is $(5, 0)$

(D) vertex of hyperbola is $(5\sqrt{3}, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two of the most elegant shapes in mathematics: the ellipse and the hyperbola. They are not just equations; they are paths, orbits, and reflections of light.

Today, we are going to bridge the gap between them. We are given an ellipse,

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

We need to construct a hyperbola that shares a deep, intimate connection with it. This is not just about solving for variables; it is about understanding how these curves interact.

### Deconstructing the Ellipse

First, let us look at our ellipse. By comparing the equation to the standard form

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

we see that $a^2 = 25$ and $b^2 = 16$. This means our semi-major axis $a$ is $5$, and our semi-minor axis $b$ is $4$.

The ellipse is stretched along the $x$-axis. Now, let us find its soul—its eccentricity. The formula $e_e = \sqrt{1 - \frac{b^2}{a^2}}$ gives us:

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

With this, we can pinpoint the foci. For a horizontal ellipse, the foci are at $(\pm ae_e, 0)$. Substituting our values, we get $(\pm 5 \cdot \frac{3}{5}, 0)$, which simplifies to $(\pm 3, 0)$. These two points are the anchors of our ellipse.

### The Bridge Between Curves

Here is where the magic happens. The problem states that our hyperbola passes through these very points, $(\pm 3, 0)$.

Because the hyperbola's axes coincide with the ellipse's axes, and it is centered at the origin, these points must be the vertices of our hyperbola. This immediately gives us the semi-transverse axis of the hyperbola: $a_h = 3$.

We have successfully built a bridge from the ellipse to the hyperbola.

### The Eccentricity Connection

We are given a fascinating condition: the product of the eccentricities is unity, $e_e \cdot e_h = 1$. Since we know $e_e = \frac{3}{5}$, it follows that $\frac{3}{5} \cdot e_h = 1$, which means $e_h = \frac{5}{3}$.

Notice how the hyperbola's eccentricity is the reciprocal of the ellipse's. This is a beautiful, symmetric relationship.

Now, we use the hyperbola's defining relation, $b_h^2 = a_h^2(e_h^2 - 1)$, to find the semi-conjugate axis. Substituting $a_h = 3$ and $e_h = \frac{5}{3}$, we get:

$$
b_h^2 = 3^2 \left( \left( \frac{5}{3} \right)^2 - 1 \right) = 9 \left( \frac{25}{9} - 1 \right) = 9 \left( \frac{16}{9} \right) = 16
$$

Everything is falling into place.

### The Final Construction

We have all the pieces of the puzzle. The standard equation for a horizontal hyperbola is

$$
\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1
$$

Plugging in our values, $a_h^2 = 9$ and $b_h^2 = 16$, we arrive at the final equation:

$$
\frac{x^2}{9} - \frac{y^2}{16} = 1
$$

To verify our results, let us find the foci of this hyperbola. The formula is $(\pm a_h e_h, 0)$. Substituting $a_h = 3$ and $e_h = \frac{5}{3}$, we get $(\pm 3 \cdot \frac{5}{3}, 0) = (\pm 5, 0)$.

We have successfully derived the equation and the foci. This journey shows that when you understand the geometric definitions, the algebra simply follows the path you have already laid out.

Select Answer:

Visualized Solution

Analyze the Ellipse

Calculate Ellipse Eccentricity $e_{e}$

Locate the Foci of the Ellipse

Determine Hyperbola Parameter $a_{h}$

Find Hyperbola Eccentricity $e_{h}$

Calculate Hyperbola Parameter $b_{h}^{2}$

Construct the Hyperbola Equation

Locate Hyperbola Foci

Conclusion and Final Check

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

Analyzing the Setup

Deconstructing the Ellipse

The Bridge Between Curves

The Eccentricity Connection

The Final Construction

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