Home
Class 12
MATHS
If the latus rectum subtends a right ang...

If the latus rectum subtends a right angle at the center of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` , then find its eccentricity.

Text Solution

Verified by Experts


(1) In the figure, latus rectum PQ subtends right angle at the centre of the hyperbola.
(2) `therefore" "OS=SP`
(3) `rArr" "ae=(b^(2))/(a)`
(4) `rArr" "e=(b^(2))/(a^(2))=e^(2)-1`
(5) `rArr" "e^(2)-e-1=0`
(6) `rArr" "e=(1+sqrt5)/(2)`
Promotional Banner

Topper's Solved these Questions

  • HYPERBOLA

    CENGAGE|Exercise Exercise 7.1|3 Videos
  • HYPERBOLA

    CENGAGE|Exercise Exercise 7.2|12 Videos
  • HIGHT AND DISTANCE

    CENGAGE|Exercise JEE Previous Year|3 Videos
  • INDEFINITE INTEGRATION

    CENGAGE|Exercise Question Bank|21 Videos

Similar Questions

Explore conceptually related problems

If it is possible to draw the tangent to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 having slope 2, then find its range of eccentricity.

The eccentricity of the hyperbola (y^(2))/(9)-(x^(2))/(25)=1 is …………………

If the latus rectum of a hyperbola forms an equilateral triangle with the vertex at the center of the hyperbola ,then find the eccentricity of the hyperbola.

Find the eccentric angles of the extremities of the latus recta of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1

Find the length of Latus rectum of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 .

Show that the acute angle between the asymptotes of the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1,(a^2> b^2), is 2cos^(-1)(1/e), where e is the eccentricity of the hyperbola.

A variable chord of the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1,(b > a), subtends a right angle at the center of the hyperbola if this chord touches. a fixed circle concentric with the hyperbola a fixed ellipse concentric with the hyperbola a fixed hyperbola concentric with the hyperbola a fixed parabola having vertex at (0, 0).

If hyperbola (x^2)/(b^2)-(y^2)/(a^2)=1 passes through the focus of ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 , then find the eccentricity of hyperbola.

Locus of perpendicular from center upon normal to the hyperbola (x^(2))/(a^(2)) -(y^(2))/(b^(2)) =1 is

Find the equation of the normal to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 at the positive end of the latus rectum.