Question 402.
Obtain the equation of conic , a focus of which lies at (2,1) , the directrix of which is x + y = 0 and which passes through (1,4) . Also identify the conic.
Answer 402.
A conic is a set of points ratio of whose distances from the focus and directrix is
a constant, e, called eccentricity.
If P(x, y) is any point on the conic,
then (x - 2)^2 + (y - 1)^2 = e^2 * (x + y)^2 / 2
As ( 1, 4) lies on it,
(1 - 2)^2 + (4 - 1)^2 = e^2 * (1 + 4)^2 /2
=> 10 = e^2 * 25/2
=> e^2 = 0.8
=> e < 1
=> conic is the ellipse.
To confirm that the conic is the ellipse, see the plot of the curve in the link below.
Wolfram Alpha Graph
Simplifying, (x - 2)^2 + (y - 1)^2 = e^2 * (x + y)^2 / 2
=> 2x^2 + 2y^2 - 8x - 4y + 10 = (0.80) (x^2 + 2xy + y^2)
=> 1.2 x^2 + 1.2 y^2 - 1.6 xy - 8x - 4y + 10 = 0
=> 3x^2 + 3y^2 - 4xy - 20x - 10y + 25 = 0
is the equation of the required conic.
Link to YA!
Obtain the equation of conic , a focus of which lies at (2,1) , the directrix of which is x + y = 0 and which passes through (1,4) . Also identify the conic.
Answer 402.
A conic is a set of points ratio of whose distances from the focus and directrix is
a constant, e, called eccentricity.
If P(x, y) is any point on the conic,
then (x - 2)^2 + (y - 1)^2 = e^2 * (x + y)^2 / 2
As ( 1, 4) lies on it,
(1 - 2)^2 + (4 - 1)^2 = e^2 * (1 + 4)^2 /2
=> 10 = e^2 * 25/2
=> e^2 = 0.8
=> e < 1
=> conic is the ellipse.
To confirm that the conic is the ellipse, see the plot of the curve in the link below.
Wolfram Alpha Graph
Simplifying, (x - 2)^2 + (y - 1)^2 = e^2 * (x + y)^2 / 2
=> 2x^2 + 2y^2 - 8x - 4y + 10 = (0.80) (x^2 + 2xy + y^2)
=> 1.2 x^2 + 1.2 y^2 - 1.6 xy - 8x - 4y + 10 = 0
=> 3x^2 + 3y^2 - 4xy - 20x - 10y + 25 = 0
is the equation of the required conic.
Link to YA!
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