If PQ is a double ordinate of the hyperbola f | vedclass

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

Question

(D) Let $P(a \sec 	heta, b 	an 	heta)$ and $Q(a \sec 	heta, -b 	an 	heta)$ be the endpoints of the double ordinate.$O(0, 0)$ is the centre of th

Vedclass · Accepted Answer

$e > 2/\sqrt{3}$

Vedclass · Answer

(D) Let $P(a \sec 	heta, b 	an 	heta)$ and $Q(a \sec 	heta, -b 	an 	heta)$ be the endpoints of the double ordinate.$O(0, 0)$ is the centre of the hyperbola.The length of the double ordinate $PQ = 2b 	an 	heta$.Since $	riangle OPQ$ is an equilateral triangle,$OP = OQ = PQ$.$OP^2 = (a \sec 	heta)^2 + (b 	an 	heta)^2 = a^2 \sec^2 	heta + b^2 	an^2 	heta$.Also,$PQ^2 = (2b 	an 	heta)^2 = 4b^2 	an^2 	heta$.Equating $OP^2 = PQ^2$,we get $a^2 \sec^2 	heta + b^2 	an^2 	heta = 4b^2 	an^2 	heta$.$a^2 \sec^2 	heta = 3b^2 	an^2 	heta$.$a^2 \frac{1}{\cos^2 	heta} = 3b^2 \frac{\sin^2 	heta}{\cos^2 	heta} \implies a^2 = 3b^2 \sin^2 	heta$.Using $b^2 = a^2(e^2 - 1)$,we have $a^2 = 3a^2(e^2 - 1) \sin^2 	heta$.$1 = 3(e^2 - 1) \sin^2 	heta \implies \sin^2 	heta = \frac{1}{3(e^2 - 1)}$.Since $0 $3(e^2 - 1) > 1 \implies e^2 - 1 > 1/3 \implies e^2 > 4/3$.Thus,$e > 2/\sqrt{3}$.

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the centre of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

Similar Questions

Suppose $A, B, C$ and $D$ are the four intersection points of the curves $\frac{x^2}{18}+\frac{y^2}{8}=1$ and $x^2-y^2=5$ in $I, II, III$ and $IV$ quadrants respectively. If $\theta_1, \theta_2, \theta_3$ and $\theta_4$ respectively are the angles between the curves at $A, B, C$ and $D$, then

The chord $PQ$ of the rectangular hyperbola $xy = a^2$ meets the $x$-axis at $A$. If $C(h, k)$ is the midpoint of $PQ$ and $O$ is the origin,then the $\Delta ACO$ is:

If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee^{\prime}$ is equal to

For $0 < \theta < \pi / 2$,if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$ is $\sqrt{7}$ times the eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$,then the value of $\theta$ is:

The area of the quadrilateral formed by the foci of the hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module