Question

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.Also reduced conic obtained to standard form and draw a rough skech of the conic obtained.

Answer 1

It is an ellipse whose equation is `3x^2-4xy+3y^2-20x-10y+25=0`

Explanation:

The distance of point `(1,4)` from focus at `(2,1)` is

`sqrt((1-2)^2+(4-1)^2)=sqrt10`

The distance of point `(1,4)` from directrix is `(1+4)/sqrt2=5/sqrt2`

and ratio of distances is `sqrt10/(5/sqrt2)=sqrt20/5=2/sqrt5<1`

As the ratio is less than `1`, it is an ellipse

and its equation is obtained from ratio of the distance of a point on ellipse say `(x,y)` from focus `(2,1)` and its distance from directrix `x+y=0` being `2/sqrt5`. The latter is `(x+y)/sqrt2`. Hence, equation is

`((x-2)^2+(y-1)^2)/((x+y)/sqrt2)^2=(2/sqrt5)^2=4/5`

or `x^2-4x+4+y^2-2y+1=2/5(x^2+2xy+y^2)`

or `5x^2-20x+20+5y^2-10y+5=2x^2+4xy+2y^2`

i.e. `3x^2-4xy+3y^2-20x-10y+25=0`

graph{(3x^2-4xy+3y^2-20x-10y+25)(x+y)((x-2)^2+(y-1)^2-0.05)((x-1)^2+(y-4)^2-0.05)=0 [-12.53, 27.47, -4.04, 15.96]}