Find the equation of the hyperbola satisfying | vedclass

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

Question

(A) The foci are $(\pm 4, 0)$,which lie on the $x$-axis. The standard equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Giv

Vedclass · Accepted Answer

$\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$

Vedclass · Answer

(A) The foci are $(\pm 4, 0)$,which lie on the $x$-axis. The standard equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Given the foci are $(\pm c, 0)$,we have $c = 4$,so $c^{2} = 16$.The length of the latus rectum is given by $\frac{2b^{2}}{a} = 12$,which implies $b^{2} = 6a$.Using the relation $c^{2} = a^{2} + b^{2}$,we substitute $c^{2} = 16$ and $b^{2} = 6a$:$a^{2} + 6a = 16$$a^{2} + 6a - 16 = 0$$(a + 8)(a - 2) = 0$Since $a > 0$,we have $a = 2$.Then $b^{2} = 6(2) = 12$.Substituting $a^{2} = 4$ and $b^{2} = 12$ into the standard equation,we get $\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$.

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 4, 0)$,the latus rectum is of length $12$.

Similar Questions

The equation $3x^2 + 7xy + 2y^2 + 5x + 5y + 2 = 0$ represents:

The equation of the tangent to the conic $x^2 - y^2 - 8x + 2y + 11 = 0$ at the point $(2, 1)$ is:

Tangents are drawn to the hyperbola $x^2 - 9y^2 = 9$ from the point $(3, 2)$. The area of the triangle formed by the tangents and the chord of contact is . . . . . . sq units.

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

$A$ hyperbola with centre at $(0,0)$ has its transverse axis along the $X$-axis,and its length is $12$. If $(8,2)$ is a point on the hyperbola,then its eccentricity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module