Question nos. 649 to 651Consider, \(E : \dfrac{(x-1)^2}{16} + \dfrac{(y-2)^2}{9} = 1\) and \(H : (x-1)^2 - (y-2)^2 = \dfrac{7}{2}\).Column-1 contains equation of tangent to either \(E\) or \(H\).Column-2 contains image of foci (whose abscissa is greater than 1) of the conic in its tangent.Column-3 contains area (in sq. units) of the triangle formed by joining foci of the conic (according to column-2), its image in the tangent and centre of the conic.Column-1Column-2Column-3(I) \(y = x + 6\)(i) \((1, \sqrt{7}+2)\)(P) \(\dfrac{7}{2}\)(II) \(y = x + 1\)(ii) \((-4, \sqrt{7}+7)\)(Q) \(\dfrac{5\sqrt{7}+7}{2}\)(III) \(x + y = 3\)(iii) \((6, \sqrt{7}-3)\)(R) \(\dfrac{7}{4}\)(IV) \(x - y - 4 = 0\)(iv) \((1, 2-\sqrt{7})\)(S) \(\dfrac{5\sqrt{7}-7}{2}\)Which of the following options is the only correct combination?
Let \(P(a\cos\theta_1, b\sin\theta_1)\), \(Q(a\cos\theta_2, b\sin\theta_2)\) and \(R(a\cos\theta_3, b\sin\theta_3)\) be the vertices of a triangle inscribed in the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). If \(\Delta_1\) = Area of \(\triangle PQR\) and \(\Delta_2\) = Area of \(\triangle P'Q'R'\) where \(P', Q', R'\) are the corresponding points on the auxiliary circle, then \(\frac{\Delta_1}{\Delta_2} = \frac{1}{7}\). Find the eccentricity of the ellipse.
If a rectangular hyperbola \((x-1)(y-2)=4\) cuts a circle \(x^2+y^2+2 g x+2 f y+c=0\) at points \((3,4),(5,3),(2,6)\) and \((-1,0)\), then the value of \((g+f)\) is equal to
For the ellipse \(\frac{x^2}{a_n^2} + \frac{y^2}{b_n^2} = 1\), the tangent at point \((x_1, y_1)\) is \(T = S_1\), i.e., \(\frac{xx_1}{a_n^2} + \frac{yy_1}{b_n^2} = \frac{x_1^2}{a_n^2} + \frac{y_1^2}{b_n^2}\). Given that \(b_n^2 x_1 = a_n^2 y_1\) and the eccentricity \(e = \frac{\sqrt{5}-1}{2}\), find the relation between \(x_1\) and \(y_1\).