If P(frac{pi}{6}) is a point on the hyperbola | vedclass

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

Question

(C) For a point $P$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the focal distances are $SP = |ex - a|$ and $S^{\prime}P = |ex + a|$.Give

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) For a point $P$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the focal distances are $SP = |ex - a|$ and $S^{\prime}P = |ex + a|$.Given the condition $SP + S^{\prime}P - 2|SP - S^{\prime}P| = 0$,we know that $|SP - S^{\prime}P| = 2a$.Substituting this into the equation: $SP + S^{\prime}P = 2(2a) = 4a$.For a hyperbola,the sum of focal distances is $SP + S^{\prime}P = 2ex$ (assuming $x > 0$).Thus,$2ex = 4a$,which implies $ex = 2a$,or $x = \frac{2a}{e}$.The point $P$ is given as $P(\frac{\pi}{6})$,which in parametric form is $(a \sec 	heta, b 	an 	heta)$.Here,$x = a \sec(\frac{\pi}{6}) = a \cdot \frac{2}{\sqrt{3}}$.Equating the two expressions for $x$: $\frac{2a}{e} = \frac{2a}{\sqrt{3}}$.Therefore,$e = \sqrt{3}$.

If $P(\frac{\pi}{6})$ is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,$S$ and $S^{\prime}$ are its foci,and $SP + S^{\prime}P - 2|SP - S^{\prime}P| = 0$,then the eccentricity $e$ is:

Similar Questions

The equation of the hyperbola with focus $(1, 2)$,eccentricity $e = \sqrt{3}$,and directrix $2x + y = 1$ is given by:

Let $x+y+1=0$ and $x-y+4=0$ be the asymptotes of a hyperbola $H$. If $(1,1)$ is a point on $H$,then the length of the latus rectum of $H$ is

If the equation of the hyperbola having $(8,3)$ and $(0,3)$ as foci and $\frac{4}{3}$ as eccentricity is $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$,then $p+q=$

The equation of the tangents to the hyperbola $3x^2 - 4y^2 = 12$ which cut equal intercepts from the axes are:

If one focus of a hyperbola is $(3,0)$,the equation of its directrix is $4x - 3y - 3 = 0$ and its eccentricity is $e = \frac{5}{4}$,then the coordinates of its vertex are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module