The equation of the hyperbola whose directrix | vedclass

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Question

(A) By the definition of a conic section,the distance from a point $P(x, y)$ to the focus $S(2, 1)$ is equal to $e$ times the perpendicular distance f

Vedclass · Accepted Answer

$x^2 - 16xy - 11y^2 - 12x + 6y + 21 = 0$

Vedclass · Answer

(A) By the definition of a conic section,the distance from a point $P(x, y)$ to the focus $S(2, 1)$ is equal to $e$ times the perpendicular distance from $P$ to the directrix $x + 2y - 1 = 0$.$SP^2 = e^2 	imes 	ext{dist}(P, 	ext{directrix})^2$$(x - 2)^2 + (y - 1)^2 = 2^2 	imes \left[ \frac{x + 2y - 1}{\sqrt{1^2 + 2^2}} ight]^2$$(x^2 - 4x + 4) + (y^2 - 2y + 1) = 4 	imes \frac{(x + 2y - 1)^2}{5}$$5(x^2 + y^2 - 4x - 2y + 5) = 4(x^2 + 4y^2 + 1 + 4xy - 2x - 4y)$$5x^2 + 5y^2 - 20x - 10y + 25 = 4x^2 + 16y^2 + 4 + 16xy - 8x - 16y$$x^2 - 16xy - 11y^2 - 12x + 6y + 21 = 0$.

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

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