The given parabola equation is y² = 4x. Comparing this with the standard form of a parabola y² = 4ax, we find that 4a = 4, which implies a = 1. The focus S of this parabola is at (a, 0), so S = (1, 0).

The given parabola equation is y² = 4x. Comparing this with the standard form of a parabola y² = 4ax, we find that 4a = 4, which implies a = 1. The focus S of this parabola is at (a, 0), so S = (1, 0).

A focal chord PQ makes an angle of 60^° with the positive x -axis. The parametric coordinates of a point P on the parabola y² = 4ax can be expressed as (atan²θ, 2atanθ) where θ is the angle the focal chord makes with the x-axis. Since a=1 and the angle is 60^°, we have P = (1 · tan²(60^°), 2 · 1 · tan(60^°)). Calculating the values, tan(60^°) = √3, so P = ((√3)², 2√3) = (3, 2√3).

A focal chord PQ makes an angle of 60^° with the positive x -axis. The parametric coordinates of a point P on the parabola y² = 4ax can be expressed as (atan²θ, 2atanθ) where θ is the angle the focal chord makes with the x-axis. Since a=1 and the angle is 60^°, we have P = (1 · tan²(60^°), 2 · 1 · tan(60^°)). Calculating the values, tan(60^°) = √3, so P = ((√3)², 2√3) = (3, 2√3).

The quadratic equation y² - 2√3y + 3 = 0 is a perfect square trinomial: (y - √3)² = 0. Solving for y, we get y = √3. Since the circle touches the y -axis at (0, α), we have α = √3. Finally, we need to find 5α². Substituting α = √3, we get 5(√3)² = 5 · 3 = 15.

The quadratic equation y² - 2√3y + 3 = 0 is a perfect square trinomial: (y - √3)² = 0. Solving for y, we get y = √3. Since the circle touches the y -axis at (0, α), we have α = √3. Finally, we need to find 5α². Substituting α = √3, we get 5(√3)² = 5 · 3 = 15.

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:

Answer: 15

Step-by-step solution

Step 1: Identify Parabola Properties and Focus

The given parabola equation is $y^2 = 4x$ . Comparing this with the standard form of a parabola $y^2 = 4ax$ , we find that $4a = 4$ , which implies $a = 1$ . The focus $S$ of this parabola is at $(a, 0)$ , so $S = (1, 0)$ .

Step 2: Determine Coordinates of P

A focal chord $PQ$ makes an angle of $60^\circ$ with the positive $x$ -axis. The parametric coordinates of a point $P$ on the parabola $y^2 = 4ax$ can be expressed as $(a\tan^2\theta, 2a\tan\theta)$ where $\theta$ is the angle the focal chord makes with the x-axis. Since $a=1$ and the angle is $60^\circ$ , we have $P = (1 \cdot \tan^2(60^\circ), 2 \cdot 1 \cdot \tan(60^\circ))$ . Calculating the values, $\tan(60^\circ) = \sqrt{3}$ , so $P = ((\sqrt{3})^2, 2\sqrt{3}) = (3, 2\sqrt{3})$ .

Step 3: Find the Equation of the Circle

The circle has $PS$ as its diameter. The coordinates of $P$ are $(3, 2\sqrt{3})$ and $S$ are $(1, 0)$ . The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ . Substituting the coordinates of $P$ and $S$ : $(x - 3)(x - 1) + (y - 2\sqrt{3})(y - 0) = 0$ . Expanding this, we get $x^2 - 4x + 3 + y^2 - 2\sqrt{3}y = 0$ , which simplifies to $x^2 + y^2 - 4x - 2\sqrt{3}y + 3 = 0$ .

Step 4: Use Tangency Condition with Y-axis

The circle touches the $y$ -axis at the point $(0, \alpha)$ . This means that when $x=0$ , the equation of the circle should yield a unique value for $y$ , which is $\alpha$ . Substitute $x=0$ into the circle's equation: $(0)^2 + y^2 - 4(0) - 2\sqrt{3}y + 3 = 0$ . This simplifies to $y^2 - 2\sqrt{3}y + 3 = 0$ .

Step 5: Solve for $\alpha$ and Calculate $5\alpha^2$

The quadratic equation $y^2 - 2\sqrt{3}y + 3 = 0$ is a perfect square trinomial: $(y - \sqrt{3})^2 = 0$ . Solving for $y$ , we get $y = \sqrt{3}$ . Since the circle touches the $y$ -axis at $(0, \alpha)$ , we have $\alpha = \sqrt{3}$ . Finally, we need to find $5\alpha^2$ . Substituting $\alpha = \sqrt{3}$ , we get $5(\sqrt{3})^2 = 5 \cdot 3 = 15$ .

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