Let $PQ$ be a focal chord of the parabola $y^{2}=4 x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment $PQ$ be also a focal chord of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{e^{2}}$ is equal to

  • [JEE MAIN 2022]
  • A

    $1+\sqrt{2}$

  • B

    $3+2 \sqrt{2}$

  • C

    $1+2 \sqrt{3}$

  • D

    $4+5 \sqrt{3}$

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An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( {\frac{1}{2},\;1} \right)$. Its one directrix is the common tangent nearer to the point $P$, to the circle ${x^2} + {y^2} = 1$ and the hyperbola ${x^2} - {y^2} = 1$. The equation of the ellipse in the standard form, is

  • [IIT 1996]

Let $'E'$  be the ellipse $\frac{{{x^2}}}{9}$$+$$\frac{{{y^2}}}{4}$ $= 1$ $\& $ $'C' $ be the circle $x^2 + y^2 = 9.$ Let $P$ $\&$ $Q$ be the points $(1 , 2) $ and $(2, 1)$  respectively. Then :

The eccentricity of an ellipse is $2/3$, latus rectum is $5$ and centre is $(0, 0)$. The equation of the ellipse is

The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length $4$ and distance between foci $4 \sqrt 2$ is-

Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $\mathrm{A}$ and $\mathrm{B}$.

$1.$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are

$(A)$ $(3,0)$ and $(0,2)$

$(B)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$(C)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$

$(D)$ $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$2.$ The orthocentre of the triangle $\mathrm{PAB}$ is

$(A)$ $\left(5, \frac{8}{7}\right)$ $(B)$ $\left(\frac{7}{5}, \frac{25}{8}\right)$

$(C)$ $\left(\frac{11}{5}, \frac{8}{5}\right)$ $(D)$ $\left(\frac{8}{25}, \frac{7}{5}\right)$

$3.$ The equation of the locus of the point whose distances from the point $\mathrm{P}$ and the line $\mathrm{AB}$ are equal, is

$(A)$ $9 x^2+y^2-6 x y-54 x-62 y+241=0$

$(B)$ $x^2+9 y^2+6 x y-54 x+62 y-241=0$

$(C)$ $9 x^2+9 y^2-6 x y-54 x-62 y-241=0$

$(D)$ $x^2+y^2-2 x y+27 x+31 y-120=0$

 Give the answer question $1,2$ and $3.$

  • [IIT 2010]