If the line x - 1 = 0, is a directrix of the hyperbola kx² - y² = 6, then the hyperbola passes throu

Q: Wait, why do we reject $k = -3$?

Great catch! Look at our standard form: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Here, $a^2 = \frac{6}{k}$. If $k$ were negative, $a^2$ would be negative, which is impossible for a real hyperbola in this standard form. We need $a^2 > 0$ for the hyperbola to exist in the real plane.

Q: Can I use the focus-directrix definition instead of the formula?

Absolutely! The definition states that the distance from any point $(x, y)$ to the focus $(ae, 0)$ is $e$ times the distance to the directrix $x = 1$. While this is the 'first principles' approach, using the standard formula $x = \frac{a}{e}$ is simply a shortcut derived from that exact definition. Both lead to the same beautiful result.

Q: Why is the directrix $x = \frac{a}{e}$ and not $x = ae$?

Remember that for a hyperbola, the eccentricity $e$ is always greater than $1$. This means $\frac{a}{e}$ is always smaller than $a$. The directrix must lie between the center and the vertex. If it were $ae$, it would be further out than the focus, which contradicts the geometry of the conic section.

Q: Is there a way to solve this without finding $k$ first?

Not really. The parameter $k$ defines the 'shape' or 'opening' of the hyperbola. Without knowing $k$, the hyperbola is just a family of curves. We must pin down the specific member of that family that satisfies the directrix condition before we can test any points.

Q: What if the directrix was $y = 1$ instead?

If the directrix were horizontal, the hyperbola would be a vertical hyperbola (transverse axis along the y-axis). The equation would then look like $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Always check the orientation of the directrix first—it tells you everything about the axis of the hyperbola!

Question

If the line $x - 1 = 0$, is a directrix of the hyperbola $kx^2 - y^2 = 6$, then the hyperbola passes through the point

Options:

(A) $(-2\sqrt{5}, 6)$

(B) $(-\sqrt{5}, 3)$

(C) $(\sqrt{5}, -2)$

(D) $(2\sqrt{5}, 3\sqrt{6})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we aren't just solving a problem; we are peeling back the layers of a conic section to reveal its hidden architecture. When you look at the equation $kx^2 - y^2 = 6$, don't just see symbols.

See a curve that is reaching out toward infinity, governed by the strict, beautiful laws of geometry. Our mission is to find the specific hyperbola that bows to the line $x = 1$ as its directrix.

### The Standard Form

Every great journey begins with a clear perspective. The equation $kx^2 - y^2 = 6$ is currently in a 'disguised' state. To understand its soul, we must bring it to the standard form.

By dividing the entire equation by $6$, we get $\frac{kx^2}{6} - \frac{y^2}{6} = 1$. Rearranging this, we see the classic structure:

$$\frac{x^2}{6/k} - \frac{y^2}{6} = 1$$

Comparing this to the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we immediately identify our parameters: $a^2 = \frac{6}{k}$ and $b^2 = 6$. This is our foundation. Without these, we are lost in the woods; with them, we have the map.

### The Directrix Constraint

Now, we encounter the gatekeeper: the directrix $x - 1 = 0$, or simply $x = 1$. In the world of hyperbolas, the directrix is not just a line; it is a boundary that dictates the eccentricity of the curve.

We know the standard formula for the directrix is $x = \frac{a}{e}$. By equating our given line to this formula, we get $\frac{a}{e} = 1$, which implies $a = e$.

Squaring both sides, we arrive at the elegant realization that $a^2 = e^2$.

### The Algebraic Bridge

We are now standing at the precipice of the solution. We have $a^2 = e^2$, and we know the fundamental identity for any hyperbola: $e^2 = 1 + \frac{b^2}{a^2}$.

By substituting our identity into this, we get $a^2 = 1 + \frac{b^2}{a^2}$. This is the moment where the physics and the algebra dance together. Substituting our values for $a^2$ and $b^2$, we obtain:

$$\frac{6}{k} = 1 + \frac{6}{6/k}$$

Simplifying the right side, we find $\frac{6}{k} = 1 + k$. Multiplying by $k$ transforms this into a beautiful quadratic equation: $k^2 + k - 6 = 0$.

### The Final Revelation

Solving the quadratic $(k + 3)(k - 2) = 0$ gives us two potential paths: $k = -3$ or $k = 2$. But wait—look back at our definition of $a^2 = \frac{6}{k}$.

If $k$ were $-3$, $a^2$ would be negative, which is physically impossible for a standard hyperbola. Thus, we reject the negative root and embrace **$k = 2$**.

Our hyperbola is officially defined: **$2x^2 - y^2 = 6$**.

To conclude, we test the point $(\sqrt{5}, -2)$. Plugging these coordinates into our equation: $2(\sqrt{5})^2 - (-2)^2 = 2(5) - 4 = 10 - 4 = 6$.

The equation holds perfectly! You have successfully navigated the constraints, identified the parameters, and verified the result. Remember, in JEE Advanced, it is not just about the answer; it is about the confidence you build by systematically dismantling the problem.

Animated Solution for Mathematics - Conic Sections: If the line $x - 1 = 0$ , is a directrix of the hyperbola $k x^{2} - y^{2} = 6$ , then the hyperbola passes through the point

Select Answer:

Visualized Solution

Standard Form of Hyperbola

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

Directrix Property

Equating Directrix

Eccentricity Formula

Substitute Values

Simplify to Quadratic

Solve for $k$

Final Hyperbola Equation

Verify the Point

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

Analyzing the Setup

The Standard Form

The Directrix Constraint

The Algebraic Bridge

The Final Revelation

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line x−1=0, is a directrix of the hyperbola kx2−y2=6, then the hyperbola passes through the point