Solution : We have, \(x^2\over 16\) – \(y^2\over 9\) = 1 Compare given equation with \(x^2\over a^2\) – \(y^2\over b^2\) = 1 a = 4 and b = 3 Since the normal to the given hyperbola whose slope is ‘m’, is y = mx \(\mp\) \({m(a^2+b^2)}\over \sqrt{a^2 – m^2b^2}\) Hence, required equation of normal is …
Find the normal to the hyperbola \(x^2\over 16\) – \(y^2\over 9\) = 1 whose slope is 1. Read More »
Solution : We have, f(x) = \(1\over x + 2\) Clearly f(x) assumes real values for all real values for all x except for the values of x satisfying x + 2 = 0 i.e. x = -2. Hence, Domain(f) = R – {-2} Similar Questions If y = 2[x] + 3 & y = …
Find the domain of the function f(x) = \(1\over x + 2\). Read More »
Solution : y = 3[x – 2] + 5 = 3[x] – 1 so 3[x] – 1 = 2[x] + 3 [x] = 4 \(\implies\) 4 \(\le\) x < 5 then y = 11 so x + y will lie in the interval [15, 16) so [x + y] = 15 Similar Questions Find the …
Solution : we have, f(x) = \(x-2\over 3-x\) Domain of f : Clearly f(x) is defined for all x satisfying 3 – x \(\ne\) 0 i.e. x \(\ne\) 3 Hence, Domain of f is R – {3} Range of f : Let y = f(x), i.e. y = \(x-2\over 3-x\) \(\implies\) 3y – xy = …
Find the domain and range of function f(x) = \(x-2\over 3-x\). Read More »
Solution : Let the equation of the ellipse be \(x^2\over a^2\) + \(y^2\over b^2\) = 1. Then, coordinates of the foci are \((\pm ae, 0)\). Therefore, ae = 2 \(\implies\) a = 4 We have \(b^2\) = \(a^2(1 – e^2)\) \(\implies\) \(b^2\) =12 Thus, the equation of the ellipse is \(x^2\over 16\) + \(y^2\over 12\) …
Solution : Let the equation of the required ellipse be \(x^2\over a^2\) + \(y^2\over b^2\) = 1 ……….(i) Since the vertices of the ellipse are on y-axis. So, the coordinates of the vertices are \((0, \pm b)\). \(\therefore\) b = 10 Now, \(a^2\) = \(b^2(1 – e^2)\) …
Solution : The given line is bx + ay – ab = 0 ………….(i) It is given that p = Length of the perpendicular from the origin to line (i) \(\implies\) p = \(|b(0) + a(0) – ab|\over {\sqrt{b^2+a^2}}\) = \(ab\over \sqrt{a^2+b^2}\) \(\implies\) \(p^2\) = \(a^2b^2\over a^2+b^2\) \(\implies\) \(1\over p^2\) = \(a^2+b^2\over a^2b^2\) \(\implies\) \(1\over …
Solution : We have line 12x – 5y + 9 = 0 and the point (2,1) Required distance = |\(12*2 – 5*1 + 9\over {\sqrt{12^2 + (-5)^2}}\)| = \(|24-5+9|\over 13\) = \(28\over 13\) Similar Questions If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = …
Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1) Read More »
Solution : Let \(S_n\) = 1 + 2 + 3 + ….. + n = \(\sum_{k=1}^{n} k\) Clearly, it is an arithmetic series with first term a = 1, common difference d = 1 and last term l = n. \(\therefore\) \(S_n\) = \(n\over 2\) (1 + n) = \(n(n+1)\over 2\) Similar Questions Let \(a_n\) …
Prove that the sum of first n natural numbers is \(n(n+1)\over 2\) Read More »
Solution : Given, \(a_2 + a_4 + a_6 + …… + a_{200}\) = \(\alpha\) ………(i) and \(a_1 + a_3 + a_5 + ….. + a_{199}\) = \(\beta\) ………(ii) On subtracting equation (ii) from equation (i), we get (\(a_2 – a_1\)) + (\(a_4 – a_3\)) + ……… + (\(a_{200} – …
