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

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

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Answer 1 · 2017-05-13T07:43:35

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

Explanation:

As $x + 4 y = 0$ $x+4y=0$ in slope intercept form is $y = - \frac{1}{4} x + 0$ $y=-1/4x+0$ , slope of the line $x + 4 y = 0$ $x+4y=0$ is $- \frac{1}{4}$ $-1/4$ .

Hence slope of tangents to the conic would also be $- \frac{1}{4}$ $-1/4$ .

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

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

$2 x + 4 x (\frac{d y}{d x}) + 4 y + 6 y \frac{d y}{d x} - 5 - 6 \frac{d y}{d x} = 0$ $2x+4x((dy)/(dx))+4y+6y(dy)/(dx)-5-6(dy)/(dx)=0$

i.e. $\frac{d y}{d x} (4 x + 6 y - 6) = - (2 x + 4 y - 5)$ $(dy)/(dx)(4x+6y-6)=-(2x+4y-5)$

and $\frac{d y}{d x} = - \frac{2 x + 4 y - 5}{4 x + 6 y - 6}$ $(dy)/(dx)=-(2x+4y-5)/(4x+6y-6)$

and as this should be $- \frac{1}{4}$ $-1/4$ we have

$- \frac{2 x + 4 y - 5}{4 x + 6 y - 6} = - \frac{1}{4}$ $-(2x+4y-5)/(4x+6y-6)=-1/4$

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

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

or $x = - \frac{5}{2} y + \frac{7}{2}$ $x=-5/2y+7/2$

Putting this in equation of conic we get

${(- \frac{5}{2} y + \frac{7}{2})}^{2} + 4 (- \frac{5}{2} y + \frac{7}{2}) y + 3 y^{2} - 5 (- \frac{5}{2} y + \frac{7}{2}) - 6 y + 3 = 0$ $(-5/2y+7/2)^2+4(-5/2y+7/2)y+3y^2-5(-5/2y+7/2)-6y+3=0$

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

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

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

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

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

and tangents are $y - 1 = - \frac{1}{4} (x - 1)$ $y-1=-1/4(x-1)$ i.e. $x + 4 y = 5$ $x+4y=5$

and $y - 3 = - \frac{1}{4} (x + 4)$ $y-3=-1/4(x+4)$ i.e. $x + 4 y = 8$ $x+4y=8$

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 · 2017-05-13T20:31:54

See below.

Explanation:

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

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

A tangent line to it with a given gradient ${\to v}_{0}$ $vec v_0$ can be written as

$L \to p = p_{0} + λ {\to v}_{0}$ $L->p=p_0+lambda vec v_0$

where $p_{0}$ $p_0$ is the tangency point.

the conic tangent space is given by

$\to t = (- f_{x}, f_{y}) = (5 - 2 x - 4 y, - 6 + 4 x + 6 y)$ $vec t = (-f_x,f_y) = (5 - 2 x - 4 y, -6 + 4 x + 6 y)$

so at the tangency point we must accomplish

$(5 - 2 x_{0} - 4 y_{0}, 4 x_{0} + 6 y_{0} - 6) = μ {\to v}_{0}$ $(5 - 2 x_0 - 4 y_0,4 x_0 + 6 y_0-6)= mu vec v_0$

here ${\to v}_{0} = (- 4, 1)$ $vec v_0 = (-4,1)$

solving for $p_{0}$ $p_0$

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

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

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

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

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