A double ordinate PQ of the hyperbola frac{x^ | vedclass

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

Question

(D) Let the coordinates of $P$ be $(x_1, y_1)$. Since $PQ$ is a double ordinate,$Q$ is $(x_1, -y_1)$.Since $\Delta OPQ$ is equilateral and $M$ is the

Vedclass · Accepted Answer

$e > \frac{2}{\sqrt{3}}$

Vedclass · Answer

(D) Let the coordinates of $P$ be $(x_1, y_1)$. Since $PQ$ is a double ordinate,$Q$ is $(x_1, -y_1)$.Since $\Delta OPQ$ is equilateral and $M$ is the midpoint of $PQ$ on the $x$-axis,$\angle POM = 30^{\circ}$.In $\Delta OMP$,$	an 30^{\circ} = \frac{PM}{OM} = \frac{y_1}{x_1} = \frac{1}{\sqrt{3}}$.Thus,$x_1 = \sqrt{3} y_1$.Since $P(x_1, y_1)$ lies on the hyperbola,$\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$.Substituting $x_1 = \sqrt{3} y_1$,we get $\frac{3y_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$,which implies $y_1^2 \left( \frac{3b^2 - a^2}{a^2b^2} ight) = 1$.For $y_1^2 > 0$,we must have $3b^2 - a^2 > 0$,so $b^2 > \frac{a^2}{3}$,or $\frac{b^2}{a^2} > \frac{1}{3}$.Using $e^2 = 1 + \frac{b^2}{a^2}$,we have $e^2 > 1 + \frac{1}{3} = \frac{4}{3}$.Therefore,$e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

$A$ double ordinate $PQ$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is such that $\Delta OPQ$ is equilateral,where $O$ is the centre of the hyperbola. Then the eccentricity $e$ satisfies the relation:

Similar Questions

The equation $13[{(x - 1)^2} + {(y - 2)^2}] = 3{(2x + 3y - 2)^2}$ represents

The locus of a point of intersection of two lines $x \sqrt{3}-y=k \sqrt{3}$ and $\sqrt{3} k x+k y=\sqrt{3}$,where $k \in R$,describes:

What is the length of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$?

The difference of the focal distances of any point on the hyperbola $9x^2 - 16y^2 = 144$ is

Consider a hyperbola $H : x^{2}-2y^{2}=4$. Let the tangent at a point $P(4, \sqrt{6})$ meet the $x$-axis at $Q$ and the latus rectum at $R(x_{1}, y_{1})$,where $x_{1}>0$. If $F$ is a focus of $H$ which is nearer to the point $P$,then the area of $\Delta QFR$ is equal to ....... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module