Let the focal chord PQ of the parabola y2=4x make an angle of 60∘ 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,α), then 5α2 is equal to:
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Step-by-Step Solution
Step 1: Identify Parabola Properties and Focus
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).
Step 2: Determine Coordinates of P
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)2,23)=(3,23).
Step 3: Find the Equation of the Circle
The circle has PS as its diameter. The coordinates of P are (3,23) and S are (1,0). The equation of a circle with diameter endpoints (x1,y1) and (x2,y2) is (x−x1)(x−x2)+(y−y1)(y−y2)=0. Substituting the coordinates of P and S: (x−3)(x−1)+(y−23)(y−0)=0. Expanding this, we get x2−4x+3+y2−23y=0, which simplifies to x2+y2−4x−23y+3=0.
Step 4: Use Tangency Condition with Y-axis
The circle touches the y -axis at the point (0,α). This means that when x=0, the equation of the circle should yield a unique value for y, which is α. Substitute x=0 into the circle's equation: (0)2+y2−4(0)−23y+3=0. This simplifies to y2−23y+3=0.
Step 5: Solve for α and Calculate 5α2
The quadratic equation y2−23y+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.