The descending order of magnitude of the ecce | vedclass

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

Question

(D) : Distance between foci is three times the distance between its directrices.$2ae = 3 	imes \frac{2a}{e}$ $\Rightarrow e^2 = 3$ $\Rightarrow e = \

Vedclass · Accepted Answer

$A, C, B$

Vedclass · Answer

(D) : Distance between foci is three times the distance between its directrices.$2ae = 3 	imes \frac{2a}{e}$ $\Rightarrow e^2 = 3$ $\Rightarrow e = \sqrt{3} \approx 1.732$$B$: The transverse axis is twice the conjugate axis.$2a = 2(2b)$ $\Rightarrow a = 2b$ $\Rightarrow b = \frac{a}{2}$.We know that $a^2e^2 = a^2 + b^2 = a^2 + \frac{a^2}{4} = \frac{5a^2}{4}$.$e^2 = \frac{5}{4} \Rightarrow e = \frac{\sqrt{5}}{2} \approx 1.118$$C$: Slopes of asymptotes are $m_1 = -1$ and $m_2 = 1$.Since $m_1 \cdot m_2 = -1$,it is a rectangular hyperbola.$e = \sqrt{2} \approx 1.414$Comparing the values: $1.732 > 1.414 > 1.118$.Thus,the descending order is $A, C, B$.

Similar Questions

Tangents are drawn to the hyperbola $x^2 - 9y^2 = 9$ from the point $(3, 2)$. The area of the triangle formed by the tangents and the chord of contact is . . . . . . sq units.

The number of integral points $(x, y)$ interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt{(x + 5\sqrt{2})^2 + y^2} - \sqrt{(x - 5\sqrt{2})^2 + y^2} = 10$ is (where an integral point $(x, y)$ means $x, y \in \mathbb{Z}$):

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

The eccentricity of the curve $x^2 - y^2 = 1$ is

If $P(\theta) = (x_1, \frac{3 \sqrt{5}}{2})$,$0 < \theta < \frac{\pi}{2}$ is a point on the hyperbola $\frac{x^2}{25} - \frac{y^2}{9} = 1$,where $\theta$ is the parameter in its parametric form,then $2 x_1 + 9 \sin^2 \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module