Area of the quadrilaterals formed by drawing tangents at the ends of latus recta of $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is
Area of the quadrilaterals formed by drawing tangents at the ends of latus recta of $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is
- A
$\frac{{16}}{{\sqrt 3 }}$
- B
$\frac{{8}}{{\sqrt 3 }}$
- C
$\frac{{4}}{{\sqrt 3 }}$
- D
$4\sqrt 3 $
Similar Questions
If the eccentricity of the two ellipse $\frac{{{x^2}}}{{169}} + \frac{{{y^2}}}{{25}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ are equal, then the value of $a/b$ is
If the eccentricity of the two ellipse $\frac{{{x^2}}}{{169}} + \frac{{{y^2}}}{{25}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ are equal, then the value of $a/b$ 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 $3.$
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]
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{4}+\frac{y^2} {25}=1$.
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{4}+\frac{y^2} {25}=1$.
Statement $-1$ : If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other, then locus of that point is always a circle
Statement $-2$ : For an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , locus of that point from which two perpendicular tangents are drawn, is $x^2 + y^2 = (a + b)^2$ .
Statement $-1$ : If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other, then locus of that point is always a circle
Statement $-2$ : For an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , locus of that point from which two perpendicular tangents are drawn, is $x^2 + y^2 = (a + b)^2$ .
The number of values of $c$ such that line $y = cx + c$, $c \in R$ touches the curve $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is
The number of values of $c$ such that line $y = cx + c$, $c \in R$ touches the curve $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is