Parametric form and focal chord:
The parabola is \(y^2 = 36x\). Its focus is \((9,0)\). A chord through the focus is a focal chord. In the standard form \(y^2 = 4ax\) with \(a=9\), points are \(\bigl(9t^2,\,18t\bigr)\). A known result: if \(P\) corresponds to parameter \(t_1\) and \(Q\) to \(t_2,\) then \(PQ\) is a focal chord iff \(t_1\,t_2 = -1.\)
Coordinates of \(P\) and \(Q\):
Let \[ P = \bigl(9t^2,\,18t\bigr), \quad Q = \Bigl(9\bigl(-\tfrac{1}{t}\bigr)^2,\,18\bigl(-\tfrac{1}{t}\bigr)\Bigr), \] so that \(t \cdot \bigl(-\tfrac{1}{t}\bigr) = -1\). Imposing \(PQ=100\) and the condition that \(P\) has positive ordinate, we arrive (from the given steps or direct calculation) at \[ P = (81,\,54), \quad Q = (1,\,-6). \] They do lie on \(y^2=36x\), pass through \((9,0)\), and are 100 units apart.
Point \(M\) such that \(PM:MQ = 3:1\):
Since \(M\) divides \(PQ\) in the ratio \(3:1\), it is found using the section formula: \[ M = P + \frac{3}{4}\,(Q - P) = (81,\,54) + \frac{3}{4}\,\bigl((1,\,{-6}) - (81,\,54)\bigr) = (21,\,9). \]
Line perpendicular to \(PQ\) through \(M\):
The slope of \(PQ\) is \[ \frac{54 - (-6)}{81 - 1} = \frac{60}{80} = \tfrac{3}{4}. \] Hence a line perpendicular to \(PQ\) has slope \(-\tfrac{4}{3}\). The equation of the perpendicular through \(M=(21,9)\) is: \[ y - 9 = -\tfrac{4}{3}\,\bigl(x - 21\bigr), \] so \[ y = -\tfrac{4}{3}\,x + 37. \]
Checking given points: We substitute each point \((x,y)\) into \(y = -\tfrac{4}{3}x + 37\):
\((3,33)\): \(-\tfrac{4}{3}\cdot 3 + 37 = -4 + 37 = 33.\) Matches.
\((6,29)\): \(-\tfrac{4}{3}\cdot 6 + 37 = -8 + 37 = 29.\) Matches.
\((-6,45)\): \(-\tfrac{4}{3}\cdot (-6) + 37 = 8 + 37 = 45.\) Matches.
\(\boxed{(-3,43)}\): \(-\tfrac{4}{3}\cdot (-3) + 37 = 4 + 37 = 41,\) not \(43.\)
Hence \((-3,43)\) does not lie on the required perpendicular, so it is the correct choice.
If the shortest distance of the parabola \(y^{2}=4x\) from the centre of the circle \(x² + y² - 4x - 16y + 64 = 0\) is d, then d2 is equal to:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,