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 :
We need to find the value of $5\alpha^2$ based on the properties of a parabola, its focal chord, and a circle touching the y-axis.
Parabola and Focus Identification
The given parabola is $y^2 = 4x$. Comparing this with the standard form $y^2 = 4ax$, we find $4a = 4$, which implies $a=1$. The focus of the parabola is located at $S = (a, 0) = (1, 0)$.
Focal Chord Properties
The focal chord PQ makes an angle of $60^\circ$ with the positive x-axis. The slope ($m$) of the chord is given by the tangent of the angle: $m = \tan(60^\circ) = \sqrt{3}$. The equation of the line PQ passing through the focus $S(1, 0)$ with slope $m=\sqrt{3}$ is given by the point-slope form $y - y_1 = m(x - x_1)$: $y - 0 = \sqrt{3}(x - 1)$ $y = \sqrt{3}(x - 1)$
Finding Points P and Q
To find the coordinates of the intersection points P and Q, substitute the equation of the line into the parabola's equation ($y^2 = 4x$): $(\sqrt{3}(x - 1))^2 = 4x$ $3(x - 1)^2 = 4x$ $3(x^2 - 2x + 1) = 4x$ $3x^2 - 6x + 3 = 4x$ $3x^2 - 10x + 3 = 0$ Factor the quadratic equation: $(3x - 1)(x - 3) = 0$ This gives two possible x-values: $x = 1/3$ and $x = 3$. For $x = 3$, $y = \sqrt{3}(3 - 1) = 2\sqrt{3}$. So, $P = (3, 2\sqrt{3})$. Since P lies in the first quadrant ($x>0, y>0$), this is point P. For $x = 1/3$, $y = \sqrt{3}(1/3 - 1) = \sqrt{3}(-2/3) = -2/\sqrt{3}$. So, $Q = (1/3, -2/\sqrt{3})$.
Circle Properties and Tangency
The circle has PS as its diameter, where $P = (3, 2\sqrt{3})$ and $S = (1, 0)$. The center of the circle is the midpoint of PS: Center $(h, k) = \left(\frac{3+1}{2}, \frac{2\sqrt{3}+0}{2}\right) = (2, \sqrt{3})$. The radius ($r$) is half the length of the diameter PS, or the distance from the center to S: $r^2 = (2-1)^2 + (\sqrt{3}-0)^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4$. So, the radius $r = 2$. The circle touches the y-axis at $(0, \alpha)$. For a circle with center $(h, k)$ and radius $r$, touching the y-axis means the distance from the center to the y-axis is equal to the radius, i.e., $|h|=r$. Here, $|2| = 2$, which is consistent. The point of tangency on the y-axis for a circle with center $(h, k)$ is $(0, k)$. Therefore, $\alpha = k$. From the center coordinates, we have $k = \sqrt{3}$. So, $\alpha = \sqrt{3}$.
Calculating the Final Value
We need to find the value of $5\alpha^2$. $5\alpha^2 = 5(\sqrt{3})^2$ $5\alpha^2 = 5(3)$ $5\alpha^2 = 15$.
Similar Questions
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Important Questions from Coordinate Geometry
- If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:
- Let $P(10, 2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are S and $S'$. If the length of its latus rectum is 8, then the square of the area of $\Delta PSS'$ is equal to :
- Let the locus of the mid-point of the chord through the origin O of the parabola $y^2 = 4x$ be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3:1, is :
- Let a circle of radius 4 pass through the origin O, the points $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab \neq 0$. Then the locus of the centroid of $\Delta OAB$ is a circle of radius
- Let a line L passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4} = \frac{y-1}{1} = \frac{z-1}{1}$ and $\frac{x-17}{1} = \frac{y-71}{1} = \frac{z}{0}$. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point $S(1, 0, -1)$ intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ______.
