Let the foci of the ellipse x²/16 + y²/7 = 1 and the hyperbola x²/144 - y²/α = 1/25 coincide. Then t

Q: Wait, why is the hyperbola equation $\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac{1}{25}$ not in standard form? Can't I just take $A^2 = 144$ and $B^2 = \alpha$?

That is a classic trap! The standard form requires the right-hand side to be $1$. By having $\frac{1}{25}$ on the right, the denominators $144$ and $\alpha$ are not actually $A^2$ and $B^2$. You must divide the entire equation by $\frac{1}{25}$ (or multiply by $25$) to normalize it. Always check the RHS before extracting parameters!

Q: Why do we use $A^2 + B^2 = A^2E^2$ for the hyperbola? I thought the relation was $B^2 = A^2(e^2 - 1)$.

You are absolutely correct! Both are valid. The relation $B^2 = A^2(e^2 - 1)$ expands to $B^2 = A^2e^2 - A^2$. If you rearrange this, you get $A^2 + B^2 = A^2e^2$. Using $A^2 + B^2 = A^2E^2$ is often faster because it directly gives you the square of the focal distance, saving you from calculating the eccentricity $E$ explicitly.

Q: Why do we reject the negative root when finding $\alpha$?

In this specific geometry, $\alpha$ represents a parameter in the denominator of the hyperbola equation $\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac{1}{25}$. For the hyperbola to exist in the standard form, the term $\frac{y^2}{\alpha}$ must be positive (since the standard form is $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$). If $\alpha$ were negative, the equation would change its nature entirely. Here, the math naturally leads us to a positive value, confirming our geometric assumptions.

Question

Let the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{7} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

Options:

(A) $\frac{32}{9}$

(B) $\frac{18}{5}$

(C) $\frac{27}{4}$

(D) $\frac{27}{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are witnessing a beautiful, geometric dance. We are looking at two distinct curves—an ellipse and a hyperbola—that are seemingly unrelated, yet they share a secret: they share the same foci.

In the world of conic sections, the foci are the "DNA" of the curve. They define its shape, its eccentricity, and its very existence. When two curves share these points, they are mathematically "locked" together.

### Unlocking the Ellipse

Imagine you are standing on the Cartesian plane. You see the ellipse:

$$\frac{x^2}{16} + \frac{y^2}{7} = 1$$

To understand its soul, we must find its foci. We compare this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Immediately, we identify the parameters: $a^2 = 16$ and $b^2 = 7$, which gives us $a = 4$ and $b = \sqrt{7}$.

What defines the "stretch" of this ellipse is the eccentricity, $e$. We recall the elegant formula:

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

Substituting our values, we get $e^2 = 1 - \frac{7}{16} = \frac{9}{16}$. Taking the square root, we find $e = \frac{3}{4}$.

The foci are located at $(\pm ae, 0)$. With $a = 4$ and $e = \frac{3}{4}$, the product $ae$ is simply $3$. Thus, our ellipse is anchored at $(\pm 3, 0)$. Keep this number $3$ in your mind; it is the key to the entire problem.

### The Hyperbola's Disguise

Now, turn your attention to the hyperbola:

$$\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac{1}{25}$$

Many students stumble here. They see $144$ and assume it is $A^2$. But look at the right-hand side; it is $\frac{1}{25}$, not $1$. This is a classic JEE trap.

To reveal the true nature of this hyperbola, we must normalize it. We multiply the entire equation by $25$ to force the right-hand side to become $1$:

$$\frac{x^2}{144/25} - \frac{y^2}{\alpha/25} = 1$$

Comparing this to the standard form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, we identify our parameters: $A^2 = \frac{144}{25}$ and $B^2 = \frac{\alpha}{25}$.

### The Bridge of Foci

The problem states that the foci of the hyperbola coincide with those of the ellipse. We already know the ellipse's foci are at $(\pm 3, 0)$. Therefore, the hyperbola's foci must also be at $(\pm 3, 0)$.

For a hyperbola, the foci are at $(\pm AE, 0)$. This implies that $AE = 3$, or $A^2E^2 = 9$.

Here is where the beauty of conic properties shines. For any hyperbola, there is a fundamental relationship:

$$A^2E^2 = A^2 + B^2$$

Since we know $A^2E^2 = 9$, we immediately know that $A^2 + B^2 = 9$. This is the bridge we needed. We substitute our expressions for $A^2$ and $B^2$ into this equation:

$$\frac{144}{25} + \frac{\alpha}{25} = 9$$

Multiplying by $25$ to clear the denominators, we get $144 + \alpha = 225$. A quick subtraction reveals **$\alpha = 81$**.

### The Final Calculation

We have $\alpha = 81$, which means $B^2 = \frac{81}{25}$. We also know $A^2 = \frac{144}{25}$, so $A = \sqrt{\frac{144}{25}} = \frac{12}{5}$.

The question asks for the length of the latus rectum of the hyperbola. We use the formula:

$$	ext{LR} = \frac{2B^2}{A}$$

Plugging in our values:

$$	ext{LR} = \frac{2 \cdot (81/25)}{12/5} = \frac{162}{25} \cdot \frac{5}{12}$$

After simplifying the fractions, we find the final result:

$$	ext{LR} = \frac{27}{10}$$

This is the essence of JEE Advanced mathematics—not just memorizing formulas, but seeing the connections between them. You have done well.

Animated Solution for Mathematics - Conic Sections: Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

Select Answer:

Visualized Solution

Analyze the Ellipse Equation

Calculate Ellipse Eccentricity $e$

Compute Eccentricity Value

Locate the Foci of the Ellipse

Standardize the Hyperbola Equation

Apply the Coinciding Foci Condition

Hyperbola Property $A^{2} E^{2} = A^{2} + B^{2}$

Set Up Equation for $α$

Solve for Parameter $α$

Calculate $B^{2}$ and $A$

Formula for Latus Rectum

Substitute and Simplify LR

Final Conclusion

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

Analyzing the Setup

Unlocking the Ellipse

The Hyperbola's Disguise

The Bridge of Foci

The Final Calculation

Similar Questions

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

For some $θ \in (0, \frac{π}{2})$ , if the eccentricity of the hyperbola, $x^{2} - y^{2} sec^{2} θ = 10$ is $5$ times the eccentricity of the ellipse, $x^{2} sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is :

Let the ellipse $E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1$ and the hyperbola $H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$ , then the value of $24 (e + L)$ is :

Let the foci of a hyperbola be $(1, 14)$ and $(1, - 12)$ . If it passes through the point $(1, 6),$ then the length of its latus-rectum is :

Let the eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1$ . If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :

Let the foci and length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be $(\pm 5, 0)$ and $50$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{a ^{2} b ^{2}} = 1$ equals