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 problem asks for the value of \( 5\alpha^2 \). We are given a parabola \( y^2 = 4x \) and its focal chord PQ which makes an angle of \( 60^\circ \) with the positive x-axis. A circle with diameter PS (S is the focus) touches the y-axis at \( (0, \alpha) \). Point P lies in the first quadrant.

Concept Used:

The solution involves concepts from coordinate geometry, specifically parabolas and circles.

1. Parabola \( y^2 = 4ax \):
   • The focus \( S \) is at the point \( (a, 0) \).
   • Any point on the parabola can be represented in parametric form as \( P(at^2, 2at) \).
2. Slope of a Line: The slope of a line segment joining points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \). If the line makes an angle \( \theta \) with the positive x-axis, its slope is \( m = \tan\theta \).
3. Circle with Diameter: For a circle whose diameter has endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \):
   • The center of the circle 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 length of the diameter.
4. Circle Touching an Axis: A circle with center \( (h, k) \) and radius \( r \) touches the y-axis if \( |h| = r \). The point of tangency is \( (0, k) \).

Step-by-Step Solution:

Step 1: Find the focus of the parabola and the parametric coordinates of P.

The equation of the parabola is \( y^2 = 4x \). Comparing this with the standard form \( y^2 = 4ax \), we get \( 4a = 4 \), which implies \( a = 1 \).

The focus \( S \) of the parabola is at \( (a, 0) \), so \( S \) is at \( (1, 0) \).

Let the coordinates of point P on the parabola be in parametric form: \( P(at^2, 2at) = P(t^2, 2t) \).

Step 2: Use the slope of the focal chord to find the parameter \( t \).

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

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

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

Equating the slope 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 this quadratic equation for \( t \):

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

This gives two possible values for \( t \): \( t_1 = \frac{2+4}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \sqrt{3} \) and \( t_2 = \frac{2-4}{2\sqrt{3}} = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} \).

Since P lies in the first quadrant, its y-coordinate \( (2t) \) must be positive, which means \( t > 0 \). Thus, we choose \( t = \sqrt{3} \).

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

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

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

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

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

So, the center is \( C(2, \sqrt{3}) \).

The radius \( r \) is the distance from the center to either P or S. Let's use S:

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

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

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

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

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

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

Final Computation & Result:

We are asked to find the value of \( 5\alpha^2 \).

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

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

Answer 2

Given the parabola \( y^2 = 4x \) and the geometry of the focal chord: \[ P( \sqrt{3}, 2\sqrt{3} ), S(1,0) \] The equation of the circle: \[ (x - 1)^2 + (y - 0)^2 = \left( \frac{\sqrt{3}}{2} \right)^2 \] The point where the circle touches the y-axis is found to be \( (0, \alpha) \), and substituting this into the equation gives \( 5\alpha^2 = 15 \).