Find the equation of the conic of which one focus lies at (2,1), one directrix is x+y=0 and it passes through (1,4)?Also identify the conic.

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Answer 1 · 2018-01-26T07:02:07

# 3x^2-4xy+3y^2-20x-10y+25=0#. graph{3x^2-4xy+3y^2-20x-10y+25=0 [-10, 10, -5, 5]}

To describe a Conic #S,# we need to know its (1) Focus #F,#

(2) the eqn. of its Directrix #d,# and (3) its Eccentricity #e.#

We are given #F and d,# so, it remains to determine #e.#

To this end, let us recall the following definition of #e :# which uses

the Focus-Directrix Property of Conic (FDP).

For any point #Q in S, (FQ)/(QM)=e,# where, #QM# denotes the

#bot"-distance from "Q" to "d.#

#(M" is the foot of "bot" from "Q" to "d)#.

We have, #F=F(2,1), d : x+y=0, and Q=Q(1,4)#.

Accordingly, #e=sqrt{(2-1)^2+(1-4)^2}/{|1+4|/sqrt(1^2+1^2)}#,

#:. e=(sqrt10*sqrt2)/5=2/sqrt5#.

Because #e < 1,# the conic #S# is Ellipse.

To find its eqn., consider any point #P=P(x,y) in S#.

Using FDP, we have,

#sqrt{(x-2)^2+(y-1)^2}=2/sqrt5*|x+y|/sqrt(1^2+1^2)#,

#5{(x-2)^2+(y-1)^2}=2(x+y)^2#,

#rArr S : 3x^2-4xy+3y^2-20x-10y+25=0#.