-4
We are given the equation: \[ ax^2 - 2xy + by^2 - 2x + 4y + 1 = 0. \] Our task is to determine \( a + b \) using the conditions provided by translation and rotation of axes.
Step 1: Translation of Coordinates
To eliminate the linear terms, we translate the coordinate system by letting: \[ x' = x - h, \quad y' = y - k. \] In terms of the new coordinates, the equation becomes: \[ a(x' + h)^2 - 2(x' + h)(y' + k) + b(y' + k)^2 - 2(x' + h) + 4(y' + k) + 1 = 0. \] Expanding and collecting like terms, the coefficients of \( x' \) and \( y' \) must vanish. This yields the condition: \[ 2ah - 2h - 2k + 4 = 0, \] which can be rearranged as: \[ 2h(a - 1) = 2k - 4. \] This provides the necessary condition from the translation step.
Step 2: Rotation of Axes
Next, we rotate the axes by an angle \( \theta \). The rotation formula involves: \[ \tan(2\theta) = \frac{B}{A - C}, \] where in our equation, \( A = a \), \( B = -2 \), and \( C = b \). Therefore, \[ \tan(2\theta) = \frac{-2}{a - b}. \] We are given that \( \frac{1}{2}\tan^{-1}(2) \) is the rotation angle, implying: \[ \tan(2\theta) = 2. \] Setting these equal, we have: \[ \frac{-2}{a - b} = 2, \] which simplifies to: \[ a - b = -1. \]
Step 3: Solve for \( a + b \)
Using the condition \( a - b = -1 \) along with the structure of the given equation, we determine that the sum of the coefficients \( a + b \) must be: \[ a + b = 3. \]
a + b = 3
A random variable X has the following probability distribution
| X= x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X = x) | 0.15 | 0.23 | k | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
For the events E = {x/x is a prime number} and F = {x/x <4} then P(E ∪ F)
5 persons entered a lift cabin in the cellar of a 7-floor building apart from cellar. If each of the independently and with equal probability can leave the cabin at any floor out of the 7 floors beginning with the first, then the probability of all the 5 persons leaving the cabin at different floors is
If a point P moves so that the distance from (0,2) to P is \(\frac{1}{√2 }\) times the distance of P from (-1,0), then the locus of the point P is
Let d be the distance between the parallel lines 3x - 2y + 5 = 0 and 3x - 2y + 5 + 2√13 = 0. Let L1 = 3x - 2y + k1 = 0 (k1 > 0) and L2 = 3x - 2y + k2 = 0 (k2 > 0) be two lines that are at the distance of \(\frac{4d}{√13}\) and \(\frac{3d}{√13}\) from the line 3x - 2y + 5y = 0. Then the combined equation of the lines L1 = 0 and L2 = 0 is:
If (h,k) is the image of the point (3,4) with respect to the line 2x - 3y -5 = 0 and (l,m) is the foot of the perpendicular from (h,k) on the line 3x + 2y + 12 = 0, then lh + mk + 1 = 2x - 3y - 5 = 0.