Solution : The equation x = acos\(\theta\) & y = bsin\(\theta\) together represent the parametric equation of ellipse \({x_1}^2\over a^2\) + \({y_1}^2\over b^2\) = 1, where \(\theta\) is a parameter. Note that if P(\(\theta\)) = (acos\(\theta\), bsin\(\theta\)) is on the ellipse then ; Q(\(\theta\)) = (acos\(\theta\), bsin\(\theta\)) is on auxilliary circle. A circle described on …
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 : For ellipse e = \(4\over 5\), so foci = (\(\pm\)4, 0) for hyperbola e = 2, so a = \(ae\over e\) = \(4\over 2\) = 2, b = \(2\sqrt{4-1}\) = 2\(\sqrt{3}\) Hence the equation of the hyperbola is \(x^2\over 4\) – \(y^2\over 12\) = 1 Similar Questions Find the equation of the ellipse …
Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4 \(\therefore\) mx – 2 = -1 \(\implies\) m = \(1\over 2\) Since \(3x^2+4y^2\) = 12 or \(x^2\over 4\) + \(y^2\over 3\) = 1 Comparing this with \(x^2\over a^2\) + \(y^2\over b^2\) = …
Solution : \(\because\) Equation of ellipse is \(9x^2 + 16y^2\) = 144 or \(x^2\over 16\) + \({(y-3)}^2\over 9\) = 1 comparing this with \(x^2\over a^2\) + \(y^2\over b^2\) = 1 then we get \(a^2\) = 16 and \(b^2\) = 9 and comparing the line y = x + k with y = mx + c …
For what value of k does the line y = x + k touches the ellipse \(9x^2 + 16y^2\) = 144. Read More »
Solution : Here S = (2, 3) & S’ is (-2, 3) and b = \(\sqrt{5}\) \(\implies\) SS’ = 4 = 2ae \(\implies\) ae = 2 but \(b^2\) = \(a^2(1-e^2)\) \(\implies\) 5 = \(a^2\) – 4 \(\implies\) a = 3 Hence the equation to major axis is y = 3. Centre of ellipse is midpoint …
