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.
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: