The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and x = ± 4/

Q: Wait, why is the directrix $x = \frac{a}{e}$ and not $x = ae$? I always get these mixed up!

It is a classic point of confusion! Remember that the focus is always 'further' from the center than the directrix. Since $e > 1$ for a hyperbola, $ae > a > \frac{a}{e}$. Thus, the focus is at $x = ae$ and the directrix is closer to the center at $x = \frac{a}{e}$. Keep this inequality in mind, and you will never mix them up again.

Q: How do I know the point of contact formula $(-\frac{a^2m}{c}, -\frac{b^2}{c})$? Do I have to memorize it?

While you can derive it by solving the intersection of the tangent $y = mx + c$ and the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, memorizing it is a huge time-saver for JEE Advanced. It comes from the condition of tangency $c^2 = a^2m^2 - b^2$. If you ever forget, just substitute $y = mx + c$ into the hyperbola equation and set the discriminant to zero!

Q: Why is the product of focal distances $|e^2x_0^2 - a^2|$? Where does this come from?

Great question! The focal distances of a point $(x_0, y_0)$ are $|ex_0 - a|$ and $|ex_0 + a|$. When you multiply these, you get $|(ex_0 - a)(ex_0 + a)|$, which is $|e^2x_0^2 - a^2|$. It is a beautiful property that simplifies the calculation significantly, saving you from having to calculate the distance to each focus separately using the distance formula.

Question

The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are 9 and $x = \pm \frac{4}{\sqrt{13}}$, respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$. If $m$ is the product of the focal distances of the point $(x_0, y_0)$, then $4e^2 + m$ is equal to

Sigma Percentile · Accepted Answer

### The Infinite Reach of the Hyperbola

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are exploring the elegant, sweeping curves of the hyperbola. This conic section is defined by its infinite reach, and our goal is to tame that infinity using the precision of algebra.

### Analyzing the Setup

Imagine you are standing at the center of a hyperbola. You are given two vital clues: the length of the latus rectum is $9$, and the directrices are located at $x = \pm \frac{4}{\sqrt{13}}$.

Let us translate this into the language of mathematics. The length of the latus rectum is defined as:

$$ \frac{2b^2}{a} = 9 \Rightarrow 2b^2 = 9a $$

Simultaneously, the directrix of a hyperbola is defined by $x = \pm \frac{a}{e}$. By comparing this to our given equation, we find:

$$ \frac{a}{e} = \frac{4}{\sqrt{13}} \Rightarrow a = \frac{4e}{\sqrt{13}} $$

### The Algebraic Bridge

Now, we need to connect these variables. The soul of the hyperbola lies in its fundamental relation: $b^2 = a^2(e^2 - 1)$. This equation is the bridge that allows us to move between the semi-axes and the eccentricity.

Substituting this into our latus rectum equation, we get:

$$ 2a^2(e^2 - 1) = 9a $$

Since $a$ is a length and cannot be zero, we can safely divide by $a$, leaving us with $2a(e^2 - 1) = 9$. Now, substitute our expression for $a$ ($a = \frac{4e}{\sqrt{13}}$) into this equation:

$$ 2 \left( \frac{4e}{\sqrt{13}} \right) (e^2 - 1) = 9 $$

Multiplying through by $\sqrt{13}$ and expanding, we arrive at the cubic equation:

$$ 8e^3 - 8e - 9\sqrt{13} = 0 $$

### Solving the Cubic

I know that seeing a cubic equation like $8e^3 - 8e - 9\sqrt{13} = 0$ can be daunting. But take a deep breath; in JEE problems, these equations often have elegant roots.

Let us test $e = \frac{\sqrt{13}}{2}$. Plugging this in:

$$ 8\left( \frac{13\sqrt{13}}{8} \right) - 8\left( \frac{\sqrt{13}}{2} \right) = 13\sqrt{13} - 4\sqrt{13} = 9\sqrt{13} $$

It matches perfectly! Our eccentricity is $e = \frac{\sqrt{13}}{2}$.

With $e$ in hand, finding $a$ and $b^2$ is a breeze. We calculate $a = \frac{4}{\sqrt{13}} \cdot \frac{\sqrt{13}}{2} = 2$. Consequently:

$$ b^2 = a^2(e^2 - 1) = 4\left( \frac{13}{4} - 1 \right) = 4\left( \frac{9}{4} \right) = 9 $$

Our hyperbola equation is:

$$ \frac{x^2}{4} - \frac{y^2}{9} = 1 $$

### The Tangent and the Finale

The problem introduces a tangent line: $y - \sqrt{3}x + \sqrt{3} = 0$, which we rewrite as $y = \sqrt{3}x - \sqrt{3}$. This is a line with slope $m_t = \sqrt{3}$ and intercept $c = -\sqrt{3}$.

The point of contact $(x_0, y_0)$ for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $(-\frac{a^2m}{c}, -\frac{b^2}{c})$. Substituting our values:

$$ x_0 = -\frac{4(\sqrt{3})}{-\sqrt{3}} = 4, \quad y_0 = -\frac{9}{-\sqrt{3}} = 3\sqrt{3} $$

Finally, the product of the focal distances $m$ is $|e^2x_0^2 - a^2|$. Substituting $e^2 = \frac{13}{4}$, $x_0 = 4$, and $a^2 = 4$:

$$ m = \left| \left( \frac{13}{4} \right)(16) - 4 \right| = |52 - 4| = 48 $$

The final value requested is $4e^2 + m$:

$$ 4\left( \frac{13}{4} \right) + 48 = 13 + 48 = \mathbf{61} $$

We have navigated the geometry, conquered the algebra, and arrived at the solution. Remember, every complex problem is just a series of simple steps waiting to be connected. Keep practicing, and keep falling in love with the process!

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola

Latus Rectum & Directrix

Fundamental Relation

Equation in $e$

Solving for Eccentricity

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

The Tangent Line

Point of Contact

Product of Focal Distances

Calculating $m$

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Infinite Reach of the Hyperbola

Analyzing the Setup

The Algebraic Bridge

Solving the Cubic

The Tangent and the Finale

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