The eccentricity of an ellipse having centre | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of an ellipse with centre at $(0, 0)$ and axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) The equation of an ellipse with centre at $(0, 0)$ and axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through $(4, -1)$,we have $\frac{16}{a^2} + \frac{1}{b^2} = 1$,which implies $16b^2 + a^2 = a^2b^2$ $(i)$.Since the ellipse passes through $(-2, 2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1$,which implies $4b^2 + 4a^2 = a^2b^2$ $(ii)$.Equating $(i)$ and $(ii)$,we get $16b^2 + a^2 = 4b^2 + 4a^2$.$12b^2 = 3a^2$,so $a^2 = 4b^2$.Substituting $a^2 = 4b^2$ into $(ii)$,we get $4b^2 + 4(4b^2) = (4b^2)b^2$,which simplifies to $20b^2 = 4b^4$,so $b^2 = 5$.Then $a^2 = 4(5) = 20$.Since $b^2 = a^2(1 - e^2)$,we have $5 = 20(1 - e^2)$.$1 - e^2 = \frac{5}{20} = \frac{1}{4}$.$e^2 = 1 - \frac{1}{4} = \frac{3}{4}$.Therefore,$e = \frac{\sqrt{3}}{2}$.

The eccentricity of an ellipse having centre at the origin,axes along the coordinate axes and passing through the points $(4, -1)$ and $(-2, 2)$ is

Similar Questions

The locus of a point which moves such that the sum of its distances from two fixed points is a constant,is

The lines $y=2x+\sqrt{76}$ and $2y+x=8$ touch the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$. If the point of intersection of these two lines lies on a circle whose centre coincides with the centre of that ellipse,then the equation of that circle is

For the ellipse $3x^2 + 4y^2 = 12$,the length of the latus rectum is:

Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci,then the square of the sum of the lengths of its minor and major axes is equal to $...........$.

If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$ for some $a \in R$,then the distance between the foci of the ellipse is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module