Let the product of the focal distances of the point P(4, 2√3) on the hyperbola H : x²/a² - y²/b² = 1

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

You absolutely could, but it would lead to a nightmare of radicals and squaring. The beauty of conic sections lies in their intrinsic properties. Using the focal distance formula $r = |ex \pm a|$ is the 'JEE way'—it is elegant, efficient, and minimizes the chance of calculation errors. Always look for the property before you start the brute-force calculation.

Q: Why do we reject the root $t = 24$ in the quadratic equation?

Great catch! We defined $t = 16 - a^2$. If $t = 24$, then $a^2 = 16 - 24 = -8$. Since $a^2$ represents the square of the semi-transverse axis, it must be positive. Always check the physical validity of your algebraic roots! In JEE Advanced, the math often gives you 'mathematically correct' but 'physically impossible' answers.

Q: Is there a general rule for the product of focal distances?

Yes! For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the product of focal distances for a point $(x_1, y_1)$ is $|e^2x_1^2 - a^2|$. It is a powerful tool to keep in your arsenal. Memorizing this property saves precious minutes during the exam.

Question

Let the product of the focal distances of the point $P(4, 2\sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be $q$. Then $p^{2} + q^{2}$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving a problem; we are exploring the architecture of the hyperbola. When you see a problem involving a hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and a point $P(4, 2\sqrt{3})$, do not let the variables intimidate you.

Imagine the two branches stretching out to infinity, and our point $P$ sitting comfortably on the right branch. This point is our anchor. The problem asks us to consider the focal distances—the lines connecting $P$ to the two foci, $S$ and $S'$.

These distances, $r_1$ and $r_2$, are the heartbeat of the hyperbola. We are told their product is 32. This is our starting line.

### The Power of the Focal Distance Formula

Many students immediately reach for the distance formula: $\sqrt{(x-ae)^2 + y^2}$. Stop. Breathe. In JEE Advanced, brute force is rarely the intended path.

The hyperbola has a secret: the focal distances are beautifully linear. For any point $P(x_1, y_1)$ on the hyperbola, the distances to the foci are given by $r_1 = |ex_1 - a|$ and $r_2 = |ex_1 + a|$.

When we multiply these, we get the product $P = |(ex_1 - a)(ex_1 + a)|$, which simplifies to $|e^2x_1^2 - a^2|$. This is the 'Aha!' moment. We have transformed a square-root-heavy geometry problem into a clean algebraic expression.

We know $x_1 = 4$, so our equation becomes:
$$|16e^2 - a^2| = 32$$

We have successfully reduced the problem to a relationship between the eccentricity $e$ and the semi-transverse axis $a$.

### The Bridge: Connecting $a$, $b$, and $e$

Now, we need to connect this to the hyperbola itself. Since $P(4, 2\sqrt{3})$ lies on the hyperbola, it must satisfy the equation:
$$\frac{4^2}{a^2} - \frac{(2\sqrt{3})^2}{b^2} = 1 \Rightarrow \frac{16}{a^2} - \frac{12}{b^2} = 1$$

We have two equations, but three variables: $a$, $b$, and $e$. We need a bridge. That bridge is the fundamental eccentricity relation: $b^2 = a^2(e^2 - 1)$.

If we rearrange this, we find $a^2e^2 = a^2 + b^2$. This allows us to replace $e^2$ in our focal product equation. Substituting $e^2 = \frac{a^2+b^2}{a^2}$ into $|16e^2 - a^2| = 32$, we get:
$$|16\left(\frac{a^2+b^2}{a^2}\right) - a^2| = 32$$

This simplifies to:
$$|16 + \frac{16b^2}{a^2} - a^2| = 32$$

### The Substitution Dance

From our point-on-hyperbola equation, we know $\frac{12}{b^2} = \frac{16-a^2}{a^2}$, which means $\frac{b^2}{a^2} = \frac{12}{16-a^2}$. Let us introduce a substitution to make the algebra sing. Let $t = 16 - a^2$.

This transforms our equation into:
$$|16 + 16\left(\frac{12}{t}\right) - (16 - t)| = 32$$

Notice the magic: the 16s cancel out! We are left with:
$$|t + \frac{192}{t}| = 32$$

Since $a^2 < 16$ (a condition derived from the hyperbola equation), $t$ must be positive. Thus, we can drop the absolute value: $t + \frac{192}{t} = 32$.

This is a simple quadratic: $t^2 - 32t + 192 = 0$. Factoring this, we get $(t - 8)(t - 24) = 0$. We reject $t = 24$ because it leads to a negative $a^2$. So, $t = 8$, which means $a^2 = 8$. Consequently, $b^2 = 12$.

### The Final Victory

We have conquered the variables. The problem asks for $p^2 + q^2$, where $p = 2b$ (conjugate axis) and $q = \frac{2b^2}{a}$ (latus rectum).

Thus, $p^2 = 4b^2 = 4(12) = 48$, and $q^2 = \frac{4b^4}{a^2} = \frac{4(144)}{8} = 72$.

Adding them together, $48 + 72 = 120$. We have arrived at the destination. The final answer is **120**.

Enter Numerical Value:

Visualized Solution

The Hyperbola and Point $P$

Focal Distances

Formula for Focal Distances

Product of Focal Distances

Substituting $x_{1} = 4$

Point $P$ Satisfies the Hyperbola

Eccentricity Relation

Connecting the Equations

Expressing in terms of $a^{2}$

Simplifying the Equation

Solving for $t$

Finding $a^{2}$ and $b^{2}$

Conjugate Axis and Latus Rectum

Calculating $p^{2}$ and $q^{2}$

Final Result

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

The Power of the Focal Distance Formula

The Bridge

The Substitution Dance

The Final Victory

Similar Questions

Let H be the hyperbola, whose foci are $(1 \pm 2, 0)$ and eccentricity is $2$ . Then the length of its latus rectum is

List-I

List-II

If distance between the foci of an ellipse is 6 and distance between its directrices is 12, then length of its latus rectum is

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :

Let $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1, a > 0, b > 0$ , be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (22 + 14)$ . If the eccentricity $H$ is $\frac{11}{2}$ , then value of $a^{2} + b^{2}$ is equal to ______.

Let one focus of the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be at $(10, 0)$ and the corresponding directrix be $x = \frac{9}{10}$ . If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of $H$ , then $9 (e^{2} + l)$ is equal to:

Let an ellipse with centre $(1, 0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along x-axis. If its minor axis subtends an angle $6 0^{\circ}$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to _______.

Let $0 < θ < \frac{π}{2}$ . If the eccentricity of the hyperbola $\frac{x ^{2}}{c o s ^{2} θ} - \frac{y ^{2}}{s i n ^{2} θ} = 1$ is greater than 2, then the length of its latus rectum lies in the interval :

Let the product of the focal distances of the point $(3, \frac{1}{2})$ on the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ $(a > b)$ , be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipses is