The equation of the hyperbola whose directrix | vedclass

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

Question

(A) Let $P(x, y)$ be any point on the hyperbola. The focus is $S(1, 1)$ and the directrix is $2x + y - 1 = 0$.By the definition of a conic section,$SP

Vedclass · Accepted Answer

$7x^2 + 12xy - 2y^2 - 2x + 4y - 7 = 0$

Vedclass · Answer

(A) Let $P(x, y)$ be any point on the hyperbola. The focus is $S(1, 1)$ and the directrix is $2x + y - 1 = 0$.By the definition of a conic section,$SP = e \cdot PM$,where $PM$ is the perpendicular distance from $P$ to the directrix.$SP^2 = e^2 \cdot PM^2$$(x - 1)^2 + (y - 1)^2 = 3 \cdot \left( \frac{2x + y - 1}{\sqrt{2^2 + 1^2}} \right)^2$$(x^2 - 2x + 1 + y^2 - 2y + 1) = 3 \cdot \frac{(2x + y - 1)^2}{5}$$5(x^2 + y^2 - 2x - 2y + 2) = 3(4x^2 + y^2 + 1 + 4xy - 2y - 4x)$$5x^2 + 5y^2 - 10x - 10y + 10 = 12x^2 + 3y^2 + 3 + 12xy - 6y - 12x$Rearranging the terms to one side:$7x^2 + 12xy - 2y^2 - 2x + 4y - 7 = 0$

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

Similar Questions

If the eccentricity of a hyperbola is $\frac{5}{3}$,then the eccentricity of its conjugate hyperbola is

The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4x+3y-7=0$ and $x-2y-1=0$ is

The equation $\frac{x^2}{29 - p} + \frac{y^2}{4 - p} = 1$ $(p \neq 4, 29)$ represents:

$A$ hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3x^{2} + 4y^{2} = 12$. Then this hyperbola does not pass through which of the following points?

The straight line $x + y = \sqrt{2}p$ will touch the hyperbola $4x^2 - 9y^2 = 36$,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module