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} –
Solution : Let the time taken to save Rs 11040 be (n + 3) months. for first three months, he saves Rs 200 each month. In (n + 3) months, 3 \(\times\) 200 + \(n\over 2\) { 2(240) + (n – 1) \(\times\) 40 } = 11040 \(\implies\) 600 + \(n\over 2\) {40(12+ n –
Solution : Let a be the first term and d (d \(\ne\) 0) be the common difference of the given AP, then \(T_{100}\) = a + (100 – 1)d = a + 99d \(T_{50}\) = a + (50 – 1)d = a + 49d \(T_{150}\) = a + (150 – 1)d = a + 149d
Solution : Since given hyperbola xy = 8 is rectangular hyperbola. And eccentricity of rectangular hyperbola is \(\sqrt{2}\) Angle between asymptotes of hyperbola is \(2sec^{-1}(e)\) \(\implies\) \(\theta\) = \(2sec^{-1}(\sqrt{2})\) \(\implies\) \(\theta\) = \(2sec^{-1}(sec 45)\) \(\implies\) \(\theta\) = 2(45) = 90 Similar Questions Find the normal to the hyperbola \(x^2\over 16\) – \(y^2\over 9\) = 1
Angle between asymptotes of hyperbola xy=8 is Read More »
Solution : The equation of tangent to the parabola \(y^2\) = 4ax is y = mx + \(a\over m\). Since it is drawn from point (2,3) Therefore it lies on tangent y = mx + \(a\over m\). \(\implies\) 3 = 2m + \(a\over m\) \(\implies\) 3m = 2\(m^2\) + a \(\implies\) 2\(m^2\) – 3m +
