If the equation frac{x^2}{7 - k} + frac{y^2}{ | vedclass

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

Question

(A) The standard form of a conic section is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. For this equation to represent a hyperbola,the coefficients of $x

Vedclass · Accepted Answer

$5 < k < 7$

Vedclass · Answer

(A) The standard form of a conic section is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. For this equation to represent a hyperbola,the coefficients of $x^2$ and $y^2$ must have opposite signs. Let $A = \frac{1}{7 - k}$ and $B = \frac{1}{5 - k}$. For a hyperbola,$A 	imes B $\left( \frac{1}{7 - k} ight) \left( \frac{1}{5 - k} ight) Multiply by $-1$ to simplify the denominators: $\left( \frac{1}{k - 7} ight) \left( \frac{1}{k - 5} ight) Using the sign scheme method,the product is negative when $k$ lies between the roots $5$ and $7$. Therefore,$5 < k < 7$.

If the equation $\frac{x^2}{7 - k} + \frac{y^2}{5 - k} = 1$ represents a hyperbola,then:

Similar Questions

If $\theta$ is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity $e = \frac{2}{\sqrt{7}-\sqrt{3}}$,then $\sin \theta = $

If $(4, 0)$ and $(-4, 0)$ are the vertices and $(6, 0)$ and $(-6, 0)$ are the foci of a hyperbola,then its eccentricity is

The equation of the tangent parallel to $y - x + 5 = 0$ drawn to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ is

The foci of the hyperbola $\frac{x^2}{16} - \frac{(y - 2)^2}{9} = 1$ are:

Let $PQ$ be a chord of the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$, perpendicular to the $x$-axis such that $OPQ$ is an equilateral triangle, where $O$ is the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of the triangle $OPQ$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module