The ellipse 4x^2 + 9y^2 = 1 intersects the po | vedclass

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

Question

(A) The equation of the ellipse is $4x^2 + 9y^2 = 1$, which can be written as $\frac{x^2}{(1/2)^2} + \frac{y^2}{(1/3)^2} = 1$.Here, $a^2 = 1/4$ and $b

Vedclass · Accepted Answer

$\frac{\sqrt{5}}{18}$

Vedclass · Answer

(A) The equation of the ellipse is $4x^2 + 9y^2 = 1$, which can be written as $\frac{x^2}{(1/2)^2} + \frac{y^2}{(1/3)^2} = 1$.Here, $a^2 = 1/4$ and $b^2 = 1/9$. Since $a^2 > b^2$, the major axis is along the $X$-axis.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/9}{1/4}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.The foci $S$ and $S'$ are at $(\pm ae, 0)$, so $S = (\frac{1}{2} \cdot \frac{\sqrt{5}}{3}, 0) = (\frac{\sqrt{5}}{6}, 0)$ and $S' = (-\frac{\sqrt{5}}{6}, 0)$.The ellipse intersects the positive $Y$-axis at $A(0, b) = (0, 1/3)$.The base of $\Delta SAS'$ is the distance $SS' = \frac{\sqrt{5}}{6} - (-\frac{\sqrt{5}}{6}) = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.The height of the triangle is the $y$-coordinate of $A$, which is $1/3$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{\sqrt{5}}{3} 	imes \frac{1}{3} = \frac{\sqrt{5}}{18}$ sq. units.

The ellipse $4x^2 + 9y^2 = 1$ intersects the positive $Y$-axis at point $A$. If $S$ and $S'$ are its foci, then the area (in sq. units) of $\Delta SAS'$ is

Similar Questions

The eccentricity of the ellipse $y^{2}+4x^{2}-12x+6y+14=0$ is

An ellipse having foci at $(3, 1)$ and $(1, 1)$ passes through the point $(1, 3)$. Then its eccentricity is

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are :

Find the equation of the ellipse with the major axis along the $x-$axis and passing through the points $(4, 3)$ and $(-1, 4)$.

If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ at all points on the ellipse other than its four vertices,then the midpoints of the tangents intercepted between the coordinate axes lie on the curve:

Vedclass Test Series

Exam Paper Generator

Online Exam Module