Consider a branch of the hyperbola x^2 - 2y^2 | vedclass

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

Question

(B) The given equation is $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$.Completing the square,we get $(x^2 - 2\sqrt{2}x + 2) - 2(y^2 + 2\sqrt{2}y + 2

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}} - 1$

Vedclass · Answer

(B) The given equation is $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$.Completing the square,we get $(x^2 - 2\sqrt{2}x + 2) - 2(y^2 + 2\sqrt{2}y + 2) = 6 + 2 - 4$,which simplifies to $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 4$.Dividing by $4$,we obtain the standard form: $\frac{(x - \sqrt{2})^2}{4} - \frac{(y + \sqrt{2})^2}{2} = 1$.Here,$a^2 = 4 \Rightarrow a = 2$ and $b^2 = 2 \Rightarrow b = \sqrt{2}$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$.The vertex $A$ is $(\sqrt{2} + a, -\sqrt{2}) = (2 + \sqrt{2}, -\sqrt{2})$.The focus $C$ nearest to $A$ is $(\sqrt{2} + ae, -\sqrt{2}) = (\sqrt{2} + 2\sqrt{\frac{3}{2}}, -\sqrt{2}) = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$.The distance $AC = ae - a = a(e - 1) = 2(\sqrt{\frac{3}{2}} - 1) = \sqrt{6} - 2$.The length of the semi-latus rectum is $\frac{b^2}{a} = \frac{2}{2} = 1$.The point $B$ is $(ae + \sqrt{2}, \frac{b^2}{a} - \sqrt{2}) = (\sqrt{6} + \sqrt{2}, 1 - \sqrt{2})$.The area of the triangle $ABC$ with base $AC$ and height $BC$ is $\frac{1}{2} 	imes AC 	imes \frac{b^2}{a} = \frac{1}{2} 	imes (\sqrt{6} - 2) 	imes 1 = \frac{\sqrt{6}}{2} - 1 = \sqrt{\frac{6}{4}} - 1 = \sqrt{\frac{3}{2}} - 1$.

Consider a branch of the hyperbola $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$,then the area of the triangle $ABC$ is

Similar Questions

If $3 \sqrt{2} x - 4 y = 12$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{5}{4}$ is its eccentricity,then $a^2 - b^2 =$

If $2x - ky + 3 = 0$ and $3x - y + 1 = 0$ are conjugate lines with respect to the hyperbola $5x^2 - 6y^2 = 15$,then $k =$

The product of the perpendiculars drawn from any point on a hyperbola to its asymptotes is

If $l$ is the maximum value of $-3x^2+4x+1$ and $m$ is the minimum value of $3x^2+4x+1$,then the equation of the hyperbola having foci at $(l, 0)$ and $(7m, 0)$ and eccentricity $e=2$ is

$A$ tangent to a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ intercepts a length of unity from each of the coordinate axes. Then the point $(a, b)$ lies on the rectangular hyperbola:

Vedclass Test Series

Exam Paper Generator

Online Exam Module