Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$, $(A<B)$ have eccentricity $\dfrac{4}{5}$. Let the lengths of the latus recta of $E_1$ and $E_2$ be $l_1$ and $l_2$, respectively, such that $2l_1^2=9l_2$. If the distance between the foci of $E_1$ is 8, then the distance between the foci of $E_2$ is
For some \(\theta \in\left(0, \frac{\pi}{2}\right)\), if the eccentricity of the hyperbola, x2 - y2 sec2 \(\theta\) = 10 is \(\sqrt{5}\) times the eccentricity of the ellipse, x2 sec2\(\theta\) + y2 = 5, then the length of the latus rectum of the ellipse, is:
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 incorrect combination?