On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$ let $P$ be a point in the second quadrant such that the tangent at $\mathrm{P}$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $\mathrm{S}^{\prime}$ be the foci of the ellipse and $\mathrm{e}$ be its eccentricity. If $\mathrm{A}$ is the area of the triangle $SPS'$ then, the value of $\left(5-\mathrm{e}^{2}\right) . \mathrm{A}$ is :

  • [JEE MAIN 2021]
  • A

    $12$

  • B

    $6$

  • C

    $14$

  • D

    $24$

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Slope of common tangents of parabola $(x -1)^2 = 4(y -2)$ and ellipse ${\left( {x - 1} \right)^2} + \frac{{{{\left( {y - 2} \right)}^2}}}{2} = 1$ are $m_1$ and $m_2$ ,then $m_1^2 + m_2^2$ is equal to

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]

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  • [JEE MAIN 2017]

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Let $L$ be a common tangent line to the curves $4 x^{2}+9 y^{2}=36$ and $(2 x)^{2}+(2 y)^{2}=31$. Then the square of the slope of the line $L$ is ..... .

  • [JEE MAIN 2021]