Let one focus of the hyperbola H : x²/a² - y²/b² = 1 be at (√10, 0) and the corresponding directrix

Q: Wait, why is the directrix $x = a/e$ and not $x = e/a$? I always get these mixed up!

That is a classic point of confusion! Remember that the eccentricity $e$ for a hyperbola is always greater than $1$. Since $a$ is the distance from the center to the vertex, $a/e$ must be smaller than $a$. Geometrically, the directrix is always 'between' the center and the vertex, so $x = a/e$ makes perfect sense. If you used $e/a$, you would be placing the directrix outside the vertex, which is incorrect.

Q: Can I use the definition of a hyperbola (distance from focus = $e \times$ distance from directrix) instead of the standard formulas?

Absolutely! That is the fundamental definition of any conic section. If you take a point $(x, y)$ on the hyperbola, its distance to the focus $(\sqrt{10}, 0)$ must equal $e$ times its distance to the line $x = 9/\sqrt{10}$. While this is a powerful method, it often leads to more complex algebra. Using the standard formulas $ae$ and $a/e$ is simply a shortcut derived from that very definition, saving you time in a high-pressure exam like JEE.

Q: Why do we assume the hyperbola is centered at the origin?

Great observation! The problem gives the equation $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. This is the definition of a standard hyperbola centered at $(0,0)$. If the center were shifted, the equation would involve $(x-h)^2$ and $(y-k)^2$. Since the problem provides the standard form, we are safe to assume the center is at the origin.

Question

Let one focus of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $x = \frac{9}{\sqrt{10}}$. 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:

Options:

(A) 14

(B) 15

(C) 16

(D) 12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a hyperbola. It is a beautiful, symmetric curve, consisting of two infinite branches that mirror each other across the $y$-axis.

The equation provided,

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

tells us everything we need to know: the center is at the origin $(0,0)$, and the transverse axis lies perfectly along the $x$-axis. This is our home base.

Before we touch any algebra, visualize these branches opening outwards, anchored by the focus and defined by the directrix. This visual foundation is the secret to mastering coordinate geometry.

### The Algebraic Dance

We are given a focus at $(\sqrt{10}, 0)$ and a directrix at $x = \frac{9}{\sqrt{10}}$. In the language of conic sections, the focus is always at $(ae, 0)$ and the directrix is the line $x = \frac{a}{e}$.

This gives us two elegant equations:

$$ae = \sqrt{10}$$

$$\frac{a}{e} = \frac{9}{\sqrt{10}}$$

Now, here is where the magic happens. Many students would try to solve for $e$ first, leading to messy square roots. But look at the symmetry!

If we multiply these two equations, the eccentricity $e$ vanishes entirely:

$$(ae) 	imes \left(\frac{a}{e}ight) = \sqrt{10} 	imes \frac{9}{\sqrt{10}}$$

$$a^2 = 9$$

Just like that, the $e$ terms cancel out, and we find $a = 3$. It is a moment of pure mathematical satisfaction.

With $a=3$, we can easily find $e^2$ by substituting back into our first equation: $3e = \sqrt{10}$, so $e^2 = \frac{10}{9}$. Note that $e^2 > 1$, which confirms our shape is indeed a hyperbola.

### The Bridge to the Latus Rectum

We have our $a$ and our $e$. Now, we need the latus rectum, $l$. To get there, we need the conjugate axis parameter, $b$.

The fundamental relationship for a hyperbola is $b^2 = a^2(e^2 - 1)$. This formula is the bridge connecting the transverse and conjugate axes.

Let us plug in our values:

$$b^2 = 9\left(\frac{10}{9} - 1ight)$$

$$b^2 = 9\left(\frac{1}{9}ight) = 1$$

Everything is collapsing into simple integers! With $b^2 = 1$ and $a = 3$, we can calculate the length of the latus rectum, $l$, using the standard formula $l = \frac{2b^2}{a}$:

$$l = \frac{2(1)}{3} = \frac{2}{3}$$

### The Final Triumph

We have reached the final stage. The problem asks for the value of $9(e^2 + l)$. We have all the components ready:

$$9\left(\frac{10}{9} + \frac{2}{3}ight)$$

Instead of finding a common denominator, let us distribute the $9$ to make the arithmetic effortless:

$$9\left(\frac{10}{9}ight) + 9\left(\frac{2}{3}ight) = 10 + 6 = 16$$

And there it is: **16**. We navigated the geometry, danced through the algebra, and arrived at the solution with precision.

Select Answer:

Visualized Solution

Standard Hyperbola

Focus of the Hyperbola

Corresponding Directrix

Setting up the Equations

Multiplying to find $a^{2}$

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

Relation for $b^{2}$

Calculating $b^{2}$

Length of Latus Rectum $l$

Calculating $l$

Setting up the Final Expression

Final Calculation

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

The Algebraic Dance

The Bridge to the Latus Rectum

The Final Triumph

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