Solution : Pair of normals are (x + 2y)(x + 3) = 0 \(\therefore\) Normals are x + 2y = 0, x + 3 = 0 Point of intersection of normals is the center of the required circle i.e. \(C_1\)(-3,3/2) and center of the given circle is \(C_2\)(2,3/2) and radius \(r^2\) = \(\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) & (\(5\over 2\), -1) i.e. y + 1 = \(7\over {5/2}\)(x – \(5\over 2\)) \(\implies\) 5y + 5 = 14x – 35 \(\implies\) 14x – 5y – 40 =
Solution : 5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 : Total number of ways : (\(5!\over 3!1!1!2!\) + \(5!\over 2!2!2!\)) \(\times\) 3! Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = \(3^7\) (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 \(4!\over 2!2!\) = 6 Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in \(3!\over 2!\) = 3 ways Therefore,
Solution : Total number of ways of dividing 48 cards(Excluding 4 Aces) in 4 groups = \(48!\over (12!)^4 4!\) Now, distribute exactly one Ace to each group of 12 cards. Total number of ways = \(48!\over (12!)^4 4!\) \(\times\) 4! Now, distribute these groups of cards among four players = \(48!\over (12!)^4 4!\) \(\times\) 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 = \(y_2 – y_1\over x_2 – x_1\). so, slope of chord passing through two points is \(4-(-3)\over 3-2\) = 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 \(y^2\) = 4ax is y = mx + \(a\over m\) Now, if it is common to both parabola, it also lies on second parabola then \(x^2\) = 4a(mx + \(a\over m\)) \(mx^2 – 4am^2 – 4a^2\) = 0 has equal roots. then its discriminant is
What is the equation of common tangent to the parabola \(y^2\) = 4ax and \(x^2\) = 4ay ? Read More »
Solution : The Equation of tangent to the parabola having slope ‘m’, is y = mx + \(a\over m\) , (m \(\ne\) 0) and point of contact is (\(a\over m^2\), \(2a\over m\)). Similar Questions The sum of the slopes of the tangent of the parabola \(y^2\)=4ax drawn from the point (2,3) is Find the locus
What is the equation of tangent to the parabola having slope m? Read More »
Question : This tricky math problem went viral a few years back after it appeared on an entrance exam in Hong Kong for six-year-olds. Supposedly the students had just 20 seconds to solve the problem! Solution : Answer is 87. This is not a maths question, its just a tricky question. If you flip the
What is the number of parking space covered by the car? Read More »
