Question

What is the equations of the tangents of the conic `x^2+4xy+3y^2-5x-6y+3=0` which are parallel to the line `x+4y=0`?

Answer 1

`x+4y=5` and `x+4y=8`

Explanation:

As `x+4y=0` in slope intercept form is `y=-1/4x+0`, slope of the line `x+4y=0` is `-1/4`.

Hence slope of tangents to the conic would also be `-1/4`.

As slope of tangent is given by first derivative, let us find it using implicit differentiation.

We have `x^2+4xy+3y^2-5x-6y+3=0` and therefore

`2x+4x((dy)/(dx))+4y+6y(dy)/(dx)-5-6(dy)/(dx)=0`

i.e. `(dy)/(dx)(4x+6y-6)=-(2x+4y-5)`

and `(dy)/(dx)=-(2x+4y-5)/(4x+6y-6)`

and as this should be `-1/4` we have

`-(2x+4y-5)/(4x+6y-6)=-1/4`

or `8x+16y-20=4x+6y-6`

or `4x+10y-14=0`

or `x=-5/2y+7/2`

Putting this in equation of conic we get

`(-5/2y+7/2)^2+4(-5/2y+7/2)y+3y^2-5(-5/2y+7/2)-6y+3=0`

or `(-5y+7)^2+8(-5y+7)y+12y^2-10(-5y+7)-24y+12=0`

or `25y^2-70y+49-40y^2+56y+12y^2+50y-70-24y+12=0`

or `-3y^2+12y-9=0` or `y^2-4y+3=0` i.e. `y=1" or "3`

and the points are `x=1" or "-4`

and points are `(1,1)` and `(-4,3)`

and tangents are `y-1=-1/4(x-1)` i.e. `x+4y=5`

and `y-3=-1/4(x+4)` i.e. `x+4y=8`

graph{(x+4y-8)(x+4y-5)(x^2+4xy+3y^2-5x-6y+3)=0 [-5.104, 4.896, -0.62, 4.38]}

Answer 2

See below.

Explanation:

Calling `p=(x,y)` we have a conic given by

`f(p)=x^2 + 4 x y + 3 y^2 - 5 x - 6 y + 3=0`

A tangent line to it with a given gradient `vec v_0` can be written as

`L->p=p_0+lambda vec v_0`

where `p_0` is the tangency point.

the conic tangent space is given by

`vec t = (-f_x,f_y) = (5 - 2 x - 4 y, -6 + 4 x + 6 y)`

so at the tangency point we must accomplish

`(5 - 2 x_0 - 4 y_0,4 x_0 + 6 y_0-6)= mu vec v_0`

here `vec v_0 = (-4,1)`

solving for `p_0`

`{(x_0 = -1/2 (3 + 10 mu)),(y_0= 2 + (7 mu)/2):}`

Now analyzing `f@L` we have

`f(p_0+lambda vec v_0) = 3/4 (1 + (2 lambda - 11 mu) (2 lambda + mu))=0`

and solving for `lambda`

`lambda = 1/2 (5 mu pm sqrt[36 mu^2-1])`

Now, the condition for tangency is `36 mu^2-1=0` or

`mu = pm 1/6`

so we have two solutions

`L->{((-2/3, 17/12)+lambda (-4,1)),((-2/3, 17/12)+lambda (-4,1)):}`