On a rectangular hyperbola x^2-y^2=a^2, a 0 | vedclass

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

Question

(B) The rectangular hyperbola is given by $x^2-y^2=a^2$. Let the coordinates of $B$ be $(a \sec 	heta, a 	an 	heta)$ and $C$ be $(a \sec 	heta, -a

Vedclass · Accepted Answer

$(2, 4]$

Vedclass · Answer

(B) The rectangular hyperbola is given by $x^2-y^2=a^2$. Let the coordinates of $B$ be $(a \sec 	heta, a 	an 	heta)$ and $C$ be $(a \sec 	heta, -a 	an 	heta)$. Since $	riangle ABC$ is equilateral,$AB^2 = BC^2$. The distance $BC = 2a 	an 	heta$. The distance $AB^2 = (a \sec 	heta - (-a))^2 + (a 	an 	heta - 0)^2 = a^2(\sec 	heta + 1)^2 + a^2 	an^2 	heta$. Equating $AB^2 = BC^2$: $a^2(\sec 	heta + 1)^2 + a^2 	an^2 	heta = (2a 	an 	heta)^2 = 4a^2 	an^2 	heta$. $(\sec 	heta + 1)^2 = 3 	an^2 	heta = 3(\sec^2 	heta - 1) = 3(\sec 	heta + 1)(\sec 	heta - 1)$. Since $\sec 	heta + 1 
eq 0$,we have $\sec 	heta + 1 = 3(\sec 	heta - 1)$. $\sec 	heta + 1 = 3 \sec 	heta - 3 \implies 2 \sec 	heta = 4 \implies \sec 	heta = 2$. Thus,$	an^2 	heta = \sec^2 	heta - 1 = 4 - 1 = 3$,so $	an 	heta = \sqrt{3}$. The side length $BC = 2a 	an 	heta = 2a \sqrt{3}$. Given the side length is $ka$,we have $ka = 2a \sqrt{3}$,so $k = 2 \sqrt{3} \approx 2 	imes 1.732 = 3.464$. Since $2 < 3.464 \leq 4$,$k$ lies in the interval $(2, 4]$.

Similar Questions

The normal to the rectangular hyperbola $xy = c^2$ at the point $t_1$ meets the curve again at the point $t_2$. Then the value of $t_1^3 t_2$ is

The product of the perpendicular distances drawn from any point on the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ to its asymptotes is

Find the equation of the hyperbola satisfying the given conditions: Vertices $(0, \pm 3)$,foci $(0, \pm 5)$.

The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is:

Two points $A$ and $B$ have coordinates $(1, 0)$ and $(-1, 0)$ respectively,and $Q$ is a point which satisfies the relation $AQ - BQ = \pm 1$. The locus of $Q$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module