The ellipse E_1: frac{x^2}{9}+frac{y^2}{4}=1 | vedclass

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

Question

(C) The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ has vertices at $(\pm 3, 0)$ and $(0, \pm 2)$.Since $E_1$ is inscribed in a rectangle $R$ with si

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ has vertices at $(\pm 3, 0)$ and $(0, \pm 2)$.Since $E_1$ is inscribed in a rectangle $R$ with sides parallel to the coordinate axes,the vertices of the rectangle $R$ are $(\pm 3, \pm 2)$.Let the equation of the ellipse $E_2$ be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.Since $E_2$ passes through $(0, 4)$,we have $\frac{0}{a^2}+\frac{16}{b^2}=1$,which gives $b^2=16$.Since $E_2$ circumscribes the rectangle $R$,it passes through the point $(3, 2)$.Substituting $(3, 2)$ into the equation of $E_2$: $\frac{9}{a^2}+\frac{4}{b^2}=1$.Substituting $b^2=16$: $\frac{9}{a^2}+\frac{4}{16}=1$ $\Rightarrow \frac{9}{a^2}+\frac{1}{4}=1$ $\Rightarrow \frac{9}{a^2}=\frac{3}{4}$ $\Rightarrow a^2=12$.For the ellipse $E_2$,$b^2 > a^2$ (since $16 > 12$),so the eccentricity $e$ is given by $a^2=b^2(1-e^2)$.$12=16(1-e^2) \Rightarrow 1-e^2=\frac{12}{16}=\frac{3}{4}$.$e^2=1-\frac{3}{4}=\frac{1}{4} \Rightarrow e=\frac{1}{2}$.

The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse $E_2$ is

Similar Questions

If lines $3x + 2y = 10$ and $-3x + 2y = 10$ are tangents at the extremities of the latus rectum of an ellipse whose centre is the origin,then the length of the latus rectum of the ellipse is:

If a normal is drawn at a variable point $P(x, y)$ on the curve $9x^2 + 16y^2 = 144$,then the maximum distance from the centre of the curve to the normal is

If $x+y+n=0, n>0$ is a normal to the ellipse $x^2+3y^2=3$ and $x+my+3=0, m < 0$ is a tangent to the ellipse $x^2+5y^2=5$,then the point of intersection of these two lines satisfies the equation

If the line $y = 4x + c$ is a tangent to the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,then $c = \dots$

The equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module