Animated Solution for Mathematics - Conic Sections: If the line $x-1=0$ is the directrix of the parabola $y^2-kx+8=0$, then one of the values of $k$ is

Unlocking the Parabola

Welcome, future engineer. Today, we are not just solving an equation; we are dissecting the anatomy of a conic section. When you look at the equation $y^2-kx+8=0$, I want you to see more than just variables and constants.

I want you to see a curve, a path, a geometric entity that is defined by its relationship to a line—the directrix. This problem is a classic JEE Advanced challenge because it tests your ability to bridge the gap between algebraic manipulation and geometric intuition.

The Anatomy of the Shift

Let us begin by looking at the equation $y^2-kx+8=0$. Your instinct might be to jump straight into formulas, but let us pause. This equation is not in the standard form $Y^2=4AX$.

It is a shifted parabola. To understand it, we must bring it home to its standard form. We rearrange the terms to isolate the $y^2$ component:

Now, we factor out the $k$. This is the moment of clarity. By writing it as $y^2=k(x-\frac{8}{k})$, we have revealed the secret of the parabola's position.

We can now clearly see that the vertex of this parabola is not at the origin $(0,0)$, but has been shifted to the point $(\frac{8}{k},0)$. This shift is the key to everything. If you miss this, the directrix calculation will be fundamentally flawed.

The Directrix—The Boundary of the Curve

Now, let us talk about the directrix. In the standard form $Y^2=4AX$, the directrix is the vertical line $X=-A$. This line is the 'boundary' that defines the parabola's curvature.

But our parabola is shifted. If the vertex is at $x_1=\frac{8}{k}$, then the entire coordinate system has shifted. The directrix, which was at $X=-A$, must now be at $x-x_1=-A$.

This gives us the equation for our directrix: $x=x_1-a$. Here, $a$ is the focal parameter, which we find by comparing our equation to $4a=k$, giving us $a=\frac{k}{4}$.

Substituting these values, we get:

This is the theoretical directrix. It is a beautiful expression because it encapsulates the entire geometry of the parabola in a single line. We are told that this line is $x=1$.

The Algebraic Bridge

We have two expressions for the same line. We equate them:

This is where many students stumble. They see fractions and panic. Do not panic. Multiply the entire equation by $4k$ to clear the denominators.

This is a standard maneuver in the JEE toolkit. Multiplying through, we get:

Rearranging this into the standard quadratic form $k^2+4k-32=0$, we are left with a simple, elegant quadratic equation. We look for two numbers that multiply to $-32$ and add to $4$.

Those numbers are $8$ and $-4$. Thus, the equation factors beautifully into $(k+8)(k-4)=0$.

Choosing the Path

We have found two solutions: $k=-8$ and $k=4$. Both are mathematically sound. Both define a parabola that satisfies the condition of having a directrix at $x=1$.

However, in the world of competitive exams, we must be pragmatic. We look at our options: $1/8,8,4,1/4$. Only $k=4$ is present.

This journey from a raw equation to a geometric understanding is what makes physics and mathematics so thrilling. You didn't just solve for $k$; you visualized the parabola, understood its shift, defined its directrix, and solved the resulting system.

Keep this mindset—always visualize, always simplify, and always trust the geometry.