If the parabola \(x^2 = 4ay\) passes through the point (2, 1), then we need to find the length of the latus rectum.
Since the point (2, 1) lies on the parabola \(x^2 = 4ay\), we can substitute x = 2 and y = 1 into the equation:
\(2^2 = 4a(1)\)
\(4 = 4a\)
\(a = 1\)
The length of the latus rectum for a parabola in the form \(x^2 = 4ay\) is given by \(4a\).
Since \(a = 1\), the length of the latus rectum is \(4(1) = 4\).
Therefore, the correct option is (B) 4.
For the parabola $ x^2 = 4ay $, the general formula for the length of the latus rectum is $ 4a $.
Given that the parabola passes through the point $ (2, 1) $, we substitute these values into the equation of the parabola: $$ x^2 = 4ay $$ Substituting $ x = 2 $ and $ y = 1 $: $$ 2^2 = 4a(1) $$ $$ 4 = 4a $$ $$ a = 1 $$ Now, the length of the latus rectum is given by $ 4a $, so: $$ \text{Length of the latus rectum} = 4(1) = 4 $$ The length of the latus rectum is 4.
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:
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2