Solution : Here S = (2, 3) & S’ is (-2, 3) and b = √5 ⟹ SS’ = 4 = 2ae ⟹ ae = 2 but b2 = a2(1−e2) ⟹ 5 = a2 – 4 ⟹ a = 3 Hence the equation to major axis is y = 3. Centre of ellipse is midpoint […]
Solution : Family of circles is x2+y2–2x–4y+1 + λ(x2+y2–1) = 0 (1 + λ)x2 + (1 + λ)y2 – 2x – 4y + (1 – λ)) = 0 \(x^2 + y^2 – {2\over {1 + \lambda}} x – {4\over {1 + \lambda}}y +
Solution : Pair of normals are (x + 2y)(x + 3) = 0 ∴ Normals are x + 2y = 0, x + 3 = 0 Point of intersection of normals is the center of the required circle i.e. C1(-3,3/2) and center of the given circle is C2(2,3/2) and radius r2 = \(\sqrt{4 + {9\over
Solution : Since the normal to the circle always passes through the center, so the equation of the normal will be the line passing through (5,6) & (52, -1) i.e. y + 1 = 75/2(x – 52) ⟹ 5y + 5 = 14x – 35 ⟹ 14x – 5y – 40 =
Find the equation of the normal to the circle x2+y2–5x+2y−48 = 0 at the point (5,6). Read More »
Solution : 5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 : Total number of ways : (5!3!1!1!2! + 5!2!2!2!) × 3! Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = 37 (as
Solution : There are 4 odd digits (1, 1, 3, 3) and 4 odd places(first, third, fifth and seventh). At these places the odd digits can be arranged in 4!2!2! = 6 Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in 3!2! = 3 ways Therefore,
Solution : Total number of ways of dividing 48 cards(Excluding 4 Aces) in 4 groups = 48!(12!)44! Now, distribute exactly one Ace to each group of 12 cards. Total number of ways = 48!(12!)44! × 4! Now, distribute these groups of cards among four players = 48!(12!)44! × 4!4!
Solution : First of all, arrange all letters of given word alphabetically : ‘ADIPR’ Total number of words starting with A _ _ _ _ = 4! = 24 Total number of words starting with D _ _ _ _ = 4! = 24 Total number of words starting with I _ _ _ _
Solution : Since, Slope of line passing through two points is m = y2–y1x2–x1. so, slope of chord passing through two points is 4−(−3)3−2 = 7 Now, Tangent line is parallel to chord. Therefore slope of tangent line is equal to slope of chord, Hence slope of tangent line
The slope of tangent parallel to the chord joining the points (2, -3) and (3, 4) is Read More »
Solution : The equation of tangent in slope form to y2 = 4ax is y = mx + am Now, if it is common to both parabola, it also lies on second parabola then x2 = 4a(mx + am) mx2–4am2–4a2 = 0 has equal roots. then its discriminant is
What is the equation of common tangent to the parabola y2 = 4ax and x2 = 4ay ? Read More »
