Let the hyperbola H : x²/a² - y²/b² = 1 pass through the point (2√2, -2√2). A parabola is drawn whos

Q: Wait, why is the directrix of the parabola $x = -ae$?

Great question! The problem states the parabola's focus is the same as the hyperbola's focus with a positive abscissa, which is $S(ae, 0)$. It also says the directrix passes through the *other* focus, $S'(-ae, 0)$. Since the focus lies on the x-axis, the directrix must be a vertical line perpendicular to the axis of the parabola. Thus, the line passing through $(-ae, 0)$ perpendicular to the x-axis is simply $x = -ae$.

Q: Why can we cancel the eccentricity $e$ in the equation $4ae = e \cdot \frac{2b^2}{a}$?

That is a very sharp observation! We can cancel $e$ because, by definition, the eccentricity $e$ of a hyperbola is always strictly greater than $1$. Since $e \neq 0$, it is perfectly safe to divide both sides by $e$ without losing any solutions. This simplification is key to unlocking the relationship between $a$ and $b$.

Q: How do I know the parabola is $y^2 = 4Ax$ and not $x^2 = 4Ay$?

Look at the focus and directrix. The focus is at $(ae, 0)$ and the directrix is $x = -ae$. Both lie on the x-axis. A parabola's axis of symmetry must pass through the focus and be perpendicular to the directrix. Since both are on the x-axis, the axis of symmetry is the x-axis itself. Therefore, the parabola must open horizontally, taking the form $y^2 = 4Ax$.

Question

Let the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ pass through the point $(2\sqrt{2}, -2\sqrt{2})$. A parabola is drawn whose focus is same as the focus of $H$ with positive abscissa and the directrix of the parabola passes through the other focus of $H$. If the length of the latus rectum of the parabola is $e$ times the length of the latus rectum of $H$, where $e$ is the eccentricity of $H$, then which of the following points lies on the parabola?

Options:

(A) $(2\sqrt{3}, 3\sqrt{2})$

(B) $(3\sqrt{3}, 6)$

(C) $(\sqrt{3}, 4\sqrt{3})$

(D) $(6\sqrt{3}, 12)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student, to a beautiful intersection of geometry and algebra. Today, we are not just solving a problem; we are witnessing a conversation between two of the most elegant curves in mathematics: the hyperbola and the parabola.

Let us peel back the layers of this problem together.

### The Hyperbola's Identity

We begin with our hyperbola, $H$, defined by the equation:

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

This is the classic, standard-form hyperbola, opening its arms to the left and right. The heartbeat of any conic section lies in its foci, which are located at $S(ae, 0)$ and $S'(-ae, 0)$.

The problem states that $H$ passes through the point $(2\sqrt{2}, -2\sqrt{2})$. Substituting these coordinates into our equation yields:

$$ \frac{8}{a^2} - \frac{8}{b^2} = 1 $$

Let us hold onto this as **Equation 1**. It is the bridge between the geometry of the hyperbola and the specific values of $a$ and $b$ that define its shape.

### The Parabola's Birth

Now, we introduce the parabola. The problem specifies that its focus is the same as the hyperbola's focus with a positive abscissa, $S(ae, 0)$.

Furthermore, its directrix passes through the other focus, $S'(-ae, 0)$. Because the focus lies on the $x$-axis, the directrix must be the vertical line $x = -ae$.

The distance between the focus $(ae, 0)$ and the directrix $x = -ae$ is exactly $2ae$. In the standard form of a parabola $y^2 = 4Ax$, the distance between the focus $(A, 0)$ and the directrix $x = -A$ is $2A$. By comparing these, we find $A = ae$.

### The Bridge (Latus Rectum)

The problem provides a fascinating relationship: the length of the latus rectum of the parabola, $LR_P$, is $e$ times the length of the latus rectum of the hyperbola, $LR_H$. We know $LR_P = 4A = 4ae$ and $LR_H = \frac{2b^2}{a}$.

Setting up the equation:

$$ 4ae = e \cdot \left( \frac{2b^2}{a} \right) $$

Since $e > 1$ for a hyperbola, we can cancel $e$ from both sides. This leaves us with $4a = \frac{2b^2}{a}$, which simplifies to:

$$ 2a^2 = b^2 $$

This is our **Equation 2**, a powerful relationship that links the hyperbola's parameters.

### Final Calculation

Now, we bring it all together. We substitute $b^2 = 2a^2$ into our first equation:

$$ \frac{8}{a^2} - \frac{8}{2a^2} = 1 $$

This simplifies to:

$$ \frac{8}{a^2} - \frac{4}{a^2} = 1 \Rightarrow \frac{4}{a^2} = 1 $$

Thus, $a^2 = 4$ and $b^2 = 8$. The eccentricity $e$ is found via:

$$ e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{8}{4} = 3 \Rightarrow e = \sqrt{3} $$

With $a = 2$ and $e = \sqrt{3}$, our parabola's parameter $A = ae = 2\sqrt{3}$. The equation of our parabola is:

$$ y^2 = 4(2\sqrt{3})x = 8\sqrt{3}x $$

Testing the point $(6\sqrt{3}, 12)$, the left side is $12^2 = 144$, and the right side is $8\sqrt{3} \cdot 6\sqrt{3} = 48 \cdot 3 = 144$. They match perfectly!

Select Answer:

Visualized Solution

Hyperbola and its Foci

Point on the Hyperbola

Focus of the Parabola

Directrix of the Parabola

Distance from Focus to Directrix

Latus Rectum Relationship

Simplifying the Condition

Solving for $a^{2}$

Finding $b^{2}$ and Eccentricity $e$

Forming the Parabola Equation

Testing the Given Options

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

Analyzing the Setup

The Hyperbola's Identity

The Parabola's Birth

The Bridge (Latus Rectum)

Final Calculation

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