Let the focal chord PQ of the parabola y² = 4x make an angle of 60° with the positive x-axis, where

Q: Wait, why do we reject the negative root for $t$?

Great observation! The problem explicitly states that point $P$ lies in the first quadrant. In the first quadrant, both the $x$-coordinate ($t^2$) and the $y$-coordinate ($2t$) must be positive. Since $t^2$ is always non-negative, the condition $2t > 0$ forces $t > 0$. Thus, we must reject the negative root.

Q: Why can't we just use the standard circle equation $(x-h)^2 + (y-k)^2 = r^2$?

You absolutely can! However, the diametric form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is much faster when you already know the endpoints of the diameter. It saves you the extra steps of finding the center $(h, k)$ and the radius $r$ separately.

Question

Let the focal chord $PQ$ of the parabola $y^2 = 4x$ make an angle of $60^\circ$ with the positive $x$-axis, where $P$ lies in the first quadrant. If the circle, whose one diameter is $PS$, $S$ being the focus of the parabola, touches the $y$-axis at the point $(0, \alpha)$, then $5\alpha^2$ is equal to :

Options:

(A) 15

(B) 25

(C) 30

(D) 20

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the elegant curve of the parabola $y^2 = 4x$. By comparing it to the standard form $y^2 = 4ax$, we immediately identify $a = 1$.

This tells us that the focus $S$ sits gracefully at $(1, 0)$. Now, consider a focal chord $PS$ passing through this focus, making a $60^\circ$ angle with the positive $x$-axis.

### The Parametric Dance

To find the coordinates of point $P$, we use the parametric form $P(t^2, 2t)$. The slope of the line $PS$ is given by $	an(60^\circ) = \sqrt{3}$.

Using the slope formula between $P(t^2, 2t)$ and $S(1, 0)$, we have:

$$\frac{2t - 0}{t^2 - 1} = \sqrt{3}$$

Cross-multiplying gives us the quadratic equation $\sqrt{3}t^2 - 2t - \sqrt{3} = 0$. Factoring this, we find $(\sqrt{3}t + 1)(t - \sqrt{3}) = 0$.

Since $P$ is in the first quadrant, $t$ must be positive, so we choose $t = \sqrt{3}$. This gives us the coordinates $P(3, 2\sqrt{3})$.

### The Circle's Anatomy

Now, we construct a circle using $PS$ as the diameter. The endpoints are $S(1, 0)$ and $P(3, 2\sqrt{3})$.

The diametric form of a circle is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Substituting our points, we get:

$$(x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0$$

This equation captures the entire circle.

### The Tangency Condition

The problem states the circle touches the $y$-axis at $(0, \alpha)$. This means the point $(0, \alpha)$ must satisfy the circle equation.

Substituting $x = 0$ and $y = \alpha$ into our circle equation, we get:

$$(0 - 1)(0 - 3) + (\alpha - 0)(\alpha - 2\sqrt{3}) = 0$$

This simplifies to $3 + \alpha^2 - 2\sqrt{3}\alpha = 0$. Notice the beauty of this expression: it is a perfect square.

We can rewrite it as $(\alpha - \sqrt{3})^2 = 0$, which implies $\alpha = \sqrt{3}$.

### The Final Victory

We have arrived at the finish line. The question asks for the value of $5\alpha^2$.

Since $\alpha = \sqrt{3}$, then $\alpha^2 = 3$. Therefore:

$$5\alpha^2 = 5 	imes 3 = \mathbf{15}$$

Through this journey, we have seen how the properties of a parabola and a circle intertwine to reveal a simple, elegant numerical result.

Select Answer:

Visualized Solution

Parabola and Focus

Parametric Coordinates of $P$

Equating the Slopes

Solving for $t$

Coordinates of Point $P$

Circle with Diameter $P S$

Substituting Endpoints

Circle Touches the $y$ -axis

Calculating $α$

Perfect Square for $α$

Final Answer: $5 α^{2}$

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

Analyzing the Setup

The Parametric Dance

The Circle's Anatomy

The Tangency Condition

The Final Victory

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Select Answer:

Visualized Solution

Parabola and Focus

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Equating the Slopes

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Circle with Diameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PS