If the points of intersection of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=4b$ (where $b > 4$) lie on the curve $y^{2}=3x^{2}$,then $b$ is equal to:

  • A
    $12$
  • B
    $5$
  • C
    $6$
  • D
    $10$

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Similar Questions

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.

Statement $I$: The equation of the directrix of the ellipse $4x^2+y^2-8x-4y+4=0$ is $3y=6-4\sqrt{3}$.
Statement $II$: The equation of the latus rectum of the ellipse $x^2+4y^2-4x-8y+4=0$ is $y=2+\sqrt{3}$.
Which of the above statement$(s)$ is (are) true?

Find the equations of the tangents to the ellipse $3x^{2} + 4y^{2} = 12$ which are perpendicular to the line $y + 2x = 4$.

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 $A$ and $B$.
$1.$ The coordinates of $A$ and $B$ 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 $PAB$ 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 $P$ and the line $AB$ 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 for questions $1, 2$ and $3.$

$A$ rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Then its dimensions are:

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