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:

• A. 15
• B. 25
• C. 30
• D. 20

Hint:

Use properties of focal chords and tangency conditions to determine points of intersection with axes.

Answer 1

The Correct Option is A

The objective is to determine the value of \( 5\alpha^2 \). Given is a parabola \( y^2 = 4x \) and its focal chord PQ, which forms a \( 60^\circ \) angle with the positive x-axis. A circle, with diameter PS (where S is the focus), is tangent to the y-axis at \( (0, \alpha) \). Point P is situated in the first quadrant.

Concepts Utilized:

The solution employs principles from coordinate geometry, specifically concerning parabolas and circles.

1. Parabola \( y^2 = 4ax \):
   • The focus \( S \) is located at \( (a, 0) \).
   • Parametric representation for any point on the parabola is \( P(at^2, 2at) \).
2. Slope of a Line: The slope of a line segment connecting \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Alternatively, if a line forms an angle \( \theta \) with the positive x-axis, its slope is \( m = \tan\theta \).
3. Circle with Diameter: For a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \):
   • The center is the midpoint of the diameter: \( C \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
   • The radius is half the diameter's length.
4. Circle Tangent to an Axis: A circle with center \( (h, k) \) and radius \( r \) is tangent to the y-axis if \( |h| = r \). The point of tangency is \( (0, k) \).

Step-by-Step Derivation:

Step 1: Determine the parabola's focus and the parametric coordinates of P.

For the parabola \( y^2 = 4x \), comparing with \( y^2 = 4ax \) yields \( 4a = 4 \), so \( a = 1 \).

The focus \( S \) is at \( (a, 0) \), thus \( S = (1, 0) \).

Let P on the parabola be represented parametrically as \( P(at^2, 2at) = P(t^2, 2t) \).

Step 2: Employ the focal chord's slope to ascertain the parameter \( t \).

The focal chord PQ passes through P and S. Its inclination with the positive x-axis is \( 60^\circ \). Consequently, the slope of chord PS is \( m = \tan(60^\circ) = \sqrt{3} \).

The slope of the line segment PS, connecting \( P(t^2, 2t) \) and \( S(1, 0) \), is:

\[ \text{Slope} = \frac{2t - 0}{t^2 - 1} = \frac{2t}{t^2 - 1} \]

Setting this slope equal to \( \sqrt{3} \):

\[ \frac{2t}{t^2 - 1} = \sqrt{3} \] \[ 2t = \sqrt{3}(t^2 - 1) \] \[ \sqrt{3}t^2 - 2t - \sqrt{3} = 0 \]

Solving the quadratic equation for \( t \):

\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} = \frac{2 \pm \sqrt{4 + 12}}{2\sqrt{3}} = \frac{2 \pm \sqrt{16}}{2\sqrt{3}} = \frac{2 \pm 4}{2\sqrt{3}} \]

The possible values for \( t \) are \( t_1 = \frac{2+4}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \sqrt{3} \) and \( t_2 = \frac{2-4}{2\sqrt{3}} = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} \).

As P is in the first quadrant, its y-coordinate (\( 2t \)) must be positive, implying \( t>0 \). Therefore, \( t = \sqrt{3} \).

The coordinates of P are \( ((\sqrt{3})^2, 2\sqrt{3}) = (3, 2\sqrt{3}) \).

Step 3: Determine the center and radius of the circle with diameter PS.

The diameter endpoints are \( P(3, 2\sqrt{3}) \) and \( S(1, 0) \).

The circle's center \( C(h, k) \) is the midpoint of PS:

\[ h = \frac{3 + 1}{2} = 2 \] \[ k = \frac{2\sqrt{3} + 0}{2} = \sqrt{3} \]

The center is thus \( C(2, \sqrt{3}) \).

The radius \( r \) is the distance from the center to either P or S. Using S:

\[ r = \sqrt{(2 - 1)^2 + (\sqrt{3} - 0)^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Step 4: Calculate the value of \( \alpha \).

The circle with center \( (h, k) = (2, \sqrt{3}) \) and radius \( r = 2 \) touches the y-axis, as \( r = |h| \) is satisfied (\( 2 = |2| \)).

The point of tangency on the y-axis is \( (0, k) \).

In this case, the tangency point is \( (0, \sqrt{3}) \).

The problem states this point is \( (0, \alpha) \), therefore \( \alpha = \sqrt{3} \).

Final Calculation & Outcome:

The value to be computed is \( 5\alpha^2 \).

\[ 5\alpha^2 = 5 (\sqrt{3})^2 = 5 \times 3 = 15 \]

The value of \( 5\alpha^2 \) is 15.