What is the equations of the tangents of the conic x^2+4xy+3y^2-5x-6y+3=0x2+4xy+3y25x6y+3=0 which are parallel to the line x+4y=0x+4y=0?

2 Answers
May 13, 2017

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

Explanation:

As x+4y=0x+4y=0 in slope intercept form is y=-1/4x+0y=14x+0, slope of the line x+4y=0x+4y=0 is -1/414.

Hence slope of tangents to the conic would also be -1/414.

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=0x2+4xy+3y25x6y+3=0 and therefore

2x+4x((dy)/(dx))+4y+6y(dy)/(dx)-5-6(dy)/(dx)=02x+4x(dydx)+4y+6ydydx56dydx=0

i.e. (dy)/(dx)(4x+6y-6)=-(2x+4y-5)dydx(4x+6y6)=(2x+4y5)

and (dy)/(dx)=-(2x+4y-5)/(4x+6y-6)dydx=2x+4y54x+6y6

and as this should be -1/414 we have

-(2x+4y-5)/(4x+6y-6)=-1/42x+4y54x+6y6=14

or 8x+16y-20=4x+6y-68x+16y20=4x+6y6

or 4x+10y-14=04x+10y14=0

or x=-5/2y+7/2x=52y+72

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(52y+72)2+4(52y+72)y+3y25(52y+72)6y+3=0

or (-5y+7)^2+8(-5y+7)y+12y^2-10(-5y+7)-24y+12=0(5y+7)2+8(5y+7)y+12y210(5y+7)24y+12=0

or 25y^2-70y+49-40y^2+56y+12y^2+50y-70-24y+12=025y270y+4940y2+56y+12y2+50y7024y+12=0

or -3y^2+12y-9=03y2+12y9=0 or y^2-4y+3=0y24y+3=0 i.e. y=1" or "3y=1 or 3

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

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

and tangents are y-1=-1/4(x-1)y1=14(x1) i.e. x+4y=5x+4y=5

and y-3=-1/4(x+4)y3=14(x+4) i.e. x+4y=8x+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]}

May 13, 2017

See below.

Explanation:

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

f(p)=x^2 + 4 x y + 3 y^2 - 5 x - 6 y + 3=0f(p)=x2+4xy+3y25x6y+3=0

A tangent line to it with a given gradient vec v_0v0 can be written as

L->p=p_0+lambda vec v_0Lp=p0+λv0

where p_0p0 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)t=(fx,fy)=(52x4y,6+4x+6y)

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(52x04y0,4x0+6y06)=μv0

here vec v_0 = (-4,1)v0=(4,1)

solving for p_0p0

{(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)):}

enter image source here