The length of the conjugate axis of a hyperbo | vedclass

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

Question

(B) Let the length of the conjugate axis be $2b$ and the length of the transverse axis be $2a$.
Given that the length of the conjugate axis is greate

Vedclass · Accepted Answer

$>\sqrt{2}$

Vedclass · Answer

(B) Let the length of the conjugate axis be $2b$ and the length of the transverse axis be $2a$.
Given that the length of the conjugate axis is greater than the length of the transverse axis, we have $2b > 2a$, which implies $b > a$.
For a hyperbola, the eccentricity $e$ is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Since $b > a$, it follows that $\frac{b^2}{a^2} > 1$.
Therefore, $e^2 = 1 + \frac{b^2}{a^2} > 1 + 1 = 2$.
Taking the square root on both sides, we get $e > \sqrt{2}$.

The length of the conjugate axis of a hyperbola is greater than the length of the transverse axis. Then, the eccentricity $e$ is

Similar Questions

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is

The locus of the foot of the perpendicular drawn from any focus to a variable tangent of the hyperbola $4x^2 - 9y^2 - 8x - 18y = 41$ is:

Let $P(4,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If the normal at $P$ intersects the $X$-axis at $(16,0)$, then the eccentricity of the hyperbola is

If the line $2x + \sqrt{6}y = 2$ touches the hyperbola $x^2 - 2y^2 = 4$,then the coordinates of the point of contact are

The value of $m$,for which the line $y = mx + \frac{25\sqrt{3}}{3}$ is a normal to the conic $\frac{x^2}{16} - \frac{y^2}{9} = 1$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module