Let H : -x²/a² + y²/b² = 1 be the hyperbola, whose eccentricity is √3 and the length of the latus re

Q: Why is the eccentricity formula $e = \sqrt{1 + \frac{a^2}{b^2}}$? I thought it was $\sqrt{1 + \frac{b^2}{a^2}}$?

Great catch! That confusion is common. For a horizontal hyperbola, the denominator under the positive term is $a^2$. For a vertical hyperbola, the positive term is the $y$ term, so the denominator is $b^2$. The eccentricity formula always involves the ratio of the square of the semi-minor axis to the square of the semi-major axis. In our vertical case, $a^2$ acts as the 'minor' component relative to the $y$-axis orientation.

Q: Can I just use the property that the product of focal distances equals $b^2$?

Be very careful! That property is often misapplied. It is safer and more robust to calculate the distances directly using the distance formula. Never rely on shortcuts unless you are certain they apply to the general case! In JEE Advanced, the most reliable path is the one derived from first principles.

Q: Why did we calculate $\alpha^2$ instead of $\alpha$?

Look at the final expression required: $\alpha^2 + \beta$. The algebra naturally leads us to $\alpha^2$. Solving for $\alpha$ would introduce unnecessary square roots and potential sign errors. In JEE Advanced, always look at the target expression first—it often dictates the most efficient path.

Question

Let $H : \frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6), \alpha > 0$ lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2 + \beta$ is equal to

Options:

(A) 172

(B) 171

(C) 169

(D) 170

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to peel back the layers of a hyperbola problem that, at first glance, might seem like a standard coordinate geometry exercise, but is actually a beautiful test of your conceptual clarity.

We begin with the equation:

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

Many students see this and immediately panic because it does not look like the standard form. However, by rearranging the terms, it becomes:

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

This is a **vertical hyperbola**, opening along the $y$-axis. This simple shift in perspective is the key to the entire problem.

### Decoding the Parameters

Now that we know the orientation, we must extract the parameters $a^2$ and $b^2$. We are given the eccentricity $e = \sqrt{3}$.

For a vertical hyperbola, the relationship is $e = \sqrt{1 + \frac{a^2}{b^2}}$. Squaring both sides gives:

$$3 = 1 + \frac{a^2}{b^2} \Rightarrow \frac{a^2}{b^2} = 2 \Rightarrow a^2 = 2b^2$$

This is our first major bridge. Next, we use the length of the latus rectum, which is given as $4\sqrt{3}$. The formula for the latus rectum of a vertical hyperbola is $\frac{2a^2}{b}$.

Setting $\frac{2a^2}{b} = 4\sqrt{3}$ and substituting $a^2 = 2b^2$, we get:

$$\frac{2(2b^2)}{b} = 4\sqrt{3} \Rightarrow 4b = 4\sqrt{3}$$

Thus, $b = \sqrt{3}$ and $b^2 = 3$. Consequently, $a^2 = 2(3) = 6$. We have successfully decoded the DNA of our hyperbola:

$$\frac{y^2}{3} - \frac{x^2}{6} = 1$$

### The Point on the Curve

We are told that the point $(\alpha, 6)$ lies on this hyperbola. This is a direct invitation to substitute.

Plugging $x = \alpha$ and $y = 6$ into our equation, we get:

$$\frac{36}{3} - \frac{\alpha^2}{6} = 1$$

This simplifies to $12 - \frac{\alpha^2}{6} = 1$, which leads us to $\frac{\alpha^2}{6} = 11$, or **$\alpha^2 = 66$**. We are halfway there.

### The Focal Distances

Finally, we must find the product of the focal distances, $\beta$. The foci of a vertical hyperbola are located at $(0, \pm be)$.

Since $b = \sqrt{3}$ and $e = \sqrt{3}$, the foci are at $(0, 3)$ and $(0, -3)$. Let our point be $P(\alpha, 6)$.

The focal distances are the distances from $P$ to these two foci. Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we find:

$$d_1 = \sqrt{(\alpha - 0)^2 + (6 - 3)^2} = \sqrt{\alpha^2 + 9}$$

$$d_2 = \sqrt{(\alpha - 0)^2 + (6 - (-3))^2} = \sqrt{\alpha^2 + 81}$$

Substituting $\alpha^2 = 66$, we get $d_1 = \sqrt{66 + 9} = \sqrt{75}$ and $d_2 = \sqrt{66 + 81} = \sqrt{147}$.

The product $\beta = d_1 	imes d_2 = \sqrt{75 	imes 147}$. Breaking these down, $\sqrt{75} = 5\sqrt{3}$ and $\sqrt{147} = 7\sqrt{3}$.

Their product is $5 	imes 7 	imes 3 = 105$. Thus, **$\beta = 105$**.

### The Final Synthesis

We have arrived at the finish line. The problem asks for $\alpha^2 + \beta$.

We found $\alpha^2 = 66$ and $\beta = 105$. Adding these together:

$$66 + 105 = 171$$

The final answer is **$171$**. Remember, in JEE Advanced, the complexity is often just a mask for fundamental definitions. Stay calm, visualize the geometry, and the math will reveal itself.

Select Answer:

Visualized Solution

Identify the Hyperbola Type

Eccentricity Formula

Relation Between $a^{2}$ and $b^{2}$

Latus Rectum Formula

Solving for $a^{2}$ and $b^{2}$

The Complete Hyperbola Equation

Finding $α^{2}$

Locating the Foci

Defining Focal Distances

Calculating Focal Distances

Finding the Product $β$

Final Calculation

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

Analyzing the Setup

Decoding the Parameters

The Point on the Curve

The Focal Distances

The Final Synthesis

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