Consider a circle with its centre lying on the focus of the parabola y² = 2px such that it touches t

Q: Wait, why did we reject the negative root $x = -\frac{3p}{2}$? Isn't it a valid solution to the quadratic?

That is a brilliant question! While $x = -\frac{3p}{2}$ is a valid algebraic solution to the quadratic equation, it fails the geometric test. Look at the parabola $y^2 = 2px$. Since $y^2$ is always non-negative, $2px$ must also be non-negative. If $p > 0$, then $x$ must be $\ge 0$. A negative $x$ would imply an imaginary $y$, which doesn't exist on the real Cartesian plane. Always check your domain!

Q: Why is the radius $p$ and not $p/2$? I thought the distance from the focus to the directrix was $a$?

Ah, let's be careful with our variables! The standard form is $y^2 = 4ax$. Here, $4a = 2p$, so $a = p/2$. The distance from the focus $(\frac{p}{2}, 0)$ to the directrix $x = -\frac{p}{2}$ is indeed $2a$. Since $2a = 2(\frac{p}{2}) = p$, the radius is $p$. Always derive the parameters from the given equation rather than memorizing standard values!

Q: Is there a faster way to solve this without full expansion?

You could potentially use the property of the latus rectum or focal chords, but for this specific problem, the algebraic substitution is the most robust and least prone to error. In JEE, sometimes the 'slow and steady' path is the fastest because it prevents silly mistakes!

Question

Consider a circle with its centre lying on the focus of the parabola $y^2 = 2px$ such that it touches the directrix of the parabola. Then a point of intersection of the circle and parabola is

Options:

(A) $(\frac{p}{2}, p)$ or $(\frac{p}{2}, -p)$

(B) $(\frac{p}{2}, \frac{p}{2})$

(C) $(-\frac{p}{2}, p)$

(D) $(-\frac{p}{2}, -\frac{p}{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at the elegant curve of a parabola defined by $y^2 = 2px$. This isn't just an equation; it is a path, a trajectory.

To truly understand it, we must first anchor ourselves by finding its heart—the focus—and its boundary—the directrix. By comparing our equation to the standard form $y^2 = 4ax$, we quickly deduce that $4a = 2p$, which means $a = \frac{p}{2}$.

Thus, the focus $S$ sits proudly at $(\frac{p}{2}, 0)$. The directrix, that silent line that defines the parabola's shape, lies at $x = -\frac{p}{2}$.

### The Circle's Identity

Now, we introduce a circle. Its center is fixed at the focus $S(\frac{p}{2}, 0)$, and it is constrained by a beautiful condition: it touches the directrix.

Geometrically, this means the radius $R$ of our circle is simply the perpendicular distance from the focus to the directrix. Calculating this distance, we find:

$$R = \left| \frac{p}{2} - \left( -\frac{p}{2} \right) \right| = p$$

With a center at $(\frac{p}{2}, 0)$ and a radius of $p$, the equation of our circle becomes:

$$(x - \frac{p}{2})^2 + y^2 = p^2$$

### The Algebraic Bridge

We are now ready to find where these two worlds—the parabola and the circle—collide. To find the intersection, we solve their equations simultaneously.

We take the parabola's $y^2 = 2px$ and substitute it into the circle's equation:

$$(x - \frac{p}{2})^2 + 2px = p^2$$

Expanding the squared term, we get:

$$x^2 - px + \frac{p^2}{4} + 2px = p^2$$

Combining like terms, we arrive at the quadratic equation:

$$x^2 + px - \frac{3p^2}{4} = 0$$

### The Final Resolution

To make the quadratic $x^2 + px - \frac{3p^2}{4} = 0$ easier to handle, we multiply by $4$ to get:

$$4x^2 + 4px - 3p^2 = 0$$

Factoring this, we find $(2x - p)(2x + 3p) = 0$, yielding $x = \frac{p}{2}$ and $x = -\frac{3p}{2}$.

But wait! We must apply our geometric intuition. Since $y^2 = 2px$ and $y^2$ cannot be negative, $x$ must be non-negative. Therefore, we reject $x = -\frac{3p}{2}$.

Substituting $x = \frac{p}{2}$ back into $y^2 = 2px$, we get $y^2 = p^2$, so $y = \pm p$. The intersection points are **$(\frac{p}{2}, p)$** and **$(\frac{p}{2}, -p)$**.

Animated Solution for Mathematics - Conic Sections: Consider a circle with its centre lying on the focus of the parabola $y^{2} = 2 p x$ such that it touches the directrix of the parabola. Then a point of intersection of the circle and parabola is

Select Answer:

Visualized Solution

Parabola $y^{2} = 2 p x$ and its Focus

Equation of the Directrix

Determining the Circle's Radius

Equation of the Circle

Setting Up the Intersection

Expanding the Equation

Simplifying to Quadratic Form

Solving for $x$

Validating the $x$ Coordinate

Determining the $y$ Coordinates

Final Intersection Points

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

Analyzing the Setup

The Circle's Identity

The Algebraic Bridge

The Final Resolution

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