The equation of the hyperbola whose directrix | vedclass

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(A) By the definition of a conic section,the distance from a point $P(x, y)$ to the focus $S(1, 1)$ is $e$ times the perpendicular distance from $P$ t

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) By the definition of a conic section,the distance from a point $P(x, y)$ to the focus $S(1, 1)$ is $e$ times the perpendicular distance from $P$ to the directrix $2x + y - 1 = 0$.The distance $PS = \sqrt{(x - 1)^2 + (y - 1)^2}$.The perpendicular distance from $P$ to the directrix is $d = \frac{|2x + y - 1|}{\sqrt{2^2 + 1^2}} = \frac{|2x + y - 1|}{\sqrt{5}}$.Using $PS^2 = e^2 d^2$ with $e = \sqrt{3}$:$(x - 1)^2 + (y - 1)^2 = 3 \cdot \frac{(2x + y - 1)^2}{5}$.$5(x^2 - 2x + 1 + y^2 - 2y + 1) = 3(4x^2 + y^2 + 1 + 4xy - 2y - 4x)$.$5x^2 + 5y^2 - 10x - 10y + 10 = 12x^2 + 3y^2 + 3 + 12xy - 6y - 12x$.Rearranging 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 $e = \sqrt{3}$ is:

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