The equation of a parabola is given by the definition: the distance from any point on the parabola to the focus is equal to the perpendicular distance from the point to the directrix.
Given:
- Focus \( F = (-2, 1) \)
- Directrix: \( 2x + y + 2 = 0 \)
The distance from the point \( P(x_1, y_1) \) on the parabola to the focus is: \[ \text{Distance to focus} = \sqrt{(x_1 + 2)^2 + (y_1 - 1)^2} \] The distance from \( P(x_1, y_1) \) to the directrix \( 2x + y + 2 = 0 \) is: \[ \text{Distance to directrix} = \frac{|2x_1 + y_1 + 2|}{\sqrt{2^2 + 1^2}} = \frac{|2x_1 + y_1 + 2|}{\sqrt{5}} \] For \( x_1 = -2 \), we substitute \( x_1 = -2 \) into both expressions: \[ \sqrt{(-2 + 2)^2 + (y_1 - 1)^2} = \frac{|2(-2) + y_1 + 2|}{\sqrt{5}} \] Simplifying both sides, we solve for \( y_1 \). After solving, we find: \[ y_1 = \frac{3}{2} \]
Thus, the sum of the ordinates of the points on the parabola is \( \frac{3}{2} \).
Equation of the parabola is given by: \[ (x + 2)^2 + (y - 1)^2 = \left(\frac{2x + y + 2}{\sqrt{5}}\right)^2 \] Multiplying both sides by 5: \[ 5[(x + 2)^2 + (y - 1)^2] = (2x + y + 2)^2 \] Substitute \(x = -2\): \[ 5(y - 1)^2 = (y - 2)^2 \] Expanding both sides: \[ 5(y^2 - 2y + 1) = y^2 - 4y + 4 \] \[ 5y^2 - 10y + 5 = y^2 - 4y + 4 \] \[ 4y^2 - 6y + 1 = 0 \] Sum of roots: \[ y_1 + y_2 = \frac{6}{4} = \frac{3}{2} \] \[ \boxed{y_1 + y_2 = \frac{3}{2}} \]
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16
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:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm. 
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 