If a directrix of a hyperbola centred at the origin and passing through the point (4, -2√3) is 5x =

Q: Wait, why is the directrix $x = \frac{a}{e}$? I always get confused between $x = \frac{a}{e}$ and $x = ae$.

That is a classic point of confusion! Remember, the focus is at $(\pm ae, 0)$, which is 'further out' from the center than the vertex $(\pm a, 0)$. Since the eccentricity $e$ for a hyperbola is always greater than $1$, dividing $a$ by $e$ makes the value smaller, pulling the directrix 'inward' towards the center, while multiplying by $e$ pushes the focus 'outward'. So, the directrix is always $x = \frac{a}{e}$.

Q: Why can't we just use the focus-directrix definition $SP = e \cdot PM$ instead of the standard equation?

You absolutely can! The definition $SP = e \cdot PM$ is the 'DNA' of the conic section. However, using it requires you to find the coordinates of the focus $(ae, 0)$ and the perpendicular distance $PM$ from the point to the directrix. While elegant, it often leads to more complex radical expressions. Using the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is usually faster for algebraic manipulation in JEE problems, but both paths lead to the same beautiful truth.

Q: I keep getting stuck on the algebra when simplifying the fractions. Is there a trick?

The trick is to avoid expanding terms too early. When you have $\frac{16}{a^2} - \frac{12}{b^2} = 1$, substitute $a^2$ first, then express $b^2$ in terms of $a^2$ and $e^2$. By keeping the terms grouped, you will often find that common factors (like $16$ or $e^2$) cancel out or simplify significantly before you have to deal with the full polynomial. Patience is your best tool here!

Question

If a directrix of a hyperbola centred at the origin and passing through the point $(4, -2\sqrt{3})$ is $5x = 4\sqrt{5}$ and its eccentricity is $e$, then :

Options:

(A) $4e^4 - 24e^2 + 35 = 0$

(B) $4e^4 + 8e^2 - 35 = 0$

(C) $4e^4 - 12e^2 - 27 = 0$

(D) $4e^4 - 24e^2 + 27 = 0$

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unraveling the Hyperbola

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a geometric entity.

We are looking at a hyperbola, a curve that represents the intersection of a plane with a double cone, a shape that embodies the balance between expansion and constraint. When you see a problem like this, do not see it as a burden of algebra. See it as a puzzle where every piece—the directrix, the point on the curve, the eccentricity—is a clue leading us to a singular, elegant truth.

### Phase 1: The Geometric Anchor

Let us begin by grounding ourselves. We are given a hyperbola centered at the origin, and we are handed a gift: the equation of its directrix, $5x = 4\sqrt{5}$.

In the language of coordinate geometry, this is our anchor. We know that for a standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the directrix is defined by the line $x = \frac{a}{e}$.

By rearranging our given equation, we find $x = \frac{4\sqrt{5}}{5}$, which simplifies beautifully to $x = \frac{4}{\sqrt{5}}$. Now, we equate our two expressions for the directrix: $\frac{a}{e} = \frac{4}{\sqrt{5}}$.

This gives us a vital bridge: $a = \frac{4e}{\sqrt{5}}$. We have successfully expressed the semi-major axis $a$ in terms of the eccentricity $e$.

### Phase 2: The Constraint of the Point

Next, we consider the point $(4, -2\sqrt{3})$. This point is not arbitrary; it is a resident of our hyperbola. It must obey the law of the curve.

Substituting these coordinates into the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get:

$$\frac{4^2}{a^2} - \frac{(-2\sqrt{3})^2}{b^2} = 1$$

This simplifies to $\frac{16}{a^2} - \frac{12}{b^2} = 1$. Here, we stand at a crossroads. We have two unknowns, $a$ and $b$, and we need to reach an equation involving only $e$.

This is where the 'soul' of the hyperbola comes into play: the relationship $b^2 = a^2(e^2 - 1)$. By substituting $a^2 = (\frac{4e}{\sqrt{5}})^2 = \frac{16e^2}{5}$, we can express $b^2$ entirely in terms of $e$ as:

$$b^2 = \frac{16e^2}{5}(e^2 - 1)$$

### Phase 3: The Algebraic Dance

Now, we enter the heart of the calculation. We must substitute our expressions for $a^2$ and $b^2$ back into our point-constraint equation.

Let us write it out:

$$\frac{16}{\frac{16e^2}{5}} - \frac{12}{\frac{16e^2(e^2-1)}{5}} = 1$$

Watch as the terms begin to simplify. The $16$ in the numerator and denominator cancels out, and the $5$ flips to the top. We are left with:

$$\frac{5}{e^2} - \frac{60}{16e^2(e^2-1)} = 1$$

Reducing the fraction $\frac{60}{16}$ to $\frac{15}{4}$, the equation becomes much friendlier:

$$\frac{5}{e^2} - \frac{15}{4e^2(e^2-1)} = 1$$

### Phase 4: The Final Convergence

We are in the home stretch. To clear the denominators, we find the common denominator, which is $4e^2(e^2-1)$. Multiplying the entire equation by this term, we get:

$$20(e^2-1) - 15 = 4e^2(e^2-1)$$

Expanding this, we arrive at $20e^2 - 20 - 15 = 4e^4 - 4e^2$. Combining the terms, we shift everything to one side to form our polynomial:

$$4e^4 - 4e^2 - 20e^2 + 35 = 0$$

Which simplifies to the final, triumphant result:

**$$4e^4 - 24e^2 + 35 = 0$$**

This matches our first option. Look at what we have achieved! You have mastered the hyperbola today. Keep this confidence, and carry it into your next challenge.

Animated Solution for Mathematics - Conic Sections: If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 23)$ is $5 x = 45$ and its eccentricity is $e$ , then :

Select Answer:

Visualized Solution

Visualizing the Hyperbola

Analyzing the Directrix

The Standard Equation

Relating $a$ , $b$ , and $e$

Substitution into the Equation

Simplifying the Fractions

Removing the Denominator

Rearranging the Equation

The Final Result

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

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

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