The focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ are

  • A
    $\frac{10}{3}, \frac{2}{3}$
  • B
    $3, 1$
  • C
    $\frac{13}{3}, \frac{5}{3}$
  • D
    $4, 2$

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If $B$ and $B^{\prime}$ are the ends of the minor axis and $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, then the area of the rhombus $SBS^{\prime}B^{\prime}$ will be

If the tangents drawn from the point $(\lambda, 3)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ are perpendicular to each other,then $\lambda = ......$

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Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis,respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ and $E_2$ at $P, Q$ and $R$,respectively. Suppose that $PQ=PR=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$,respectively,then the correct expression$(s)$ is(are):
$(A) e_1^2+e_2^2=\frac{43}{40}$
$(B) e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C) |e_1^2-e_2^2|=\frac{5}{8}$
$(D) e_1 e_2=\frac{\sqrt{3}}{4}$

$P$ is a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. When the area of $\Delta PSS'$ is maximum,the inradius of $\Delta PSS'$ ($S$ and $S'$ are foci) is equal to:

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The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is

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