Geometry or calculus? II

A very well known result will greatly help us here that is :

Combined equations of tangents from a point $(u,v)$ to a conic

$a{x}^{2}+2hxy+b{y}^{2}+2gx+2fy+c=0$

is given by :

${T}^{2}=S{S}_{1}$

where $T=aux+h(xv+yu)+bvy+g(x+u)+f(y+v)+c$

$S=a{x}^{2}+2hxy+b{y}^{2}+2gx+2fy+c$

${S}_{1}=a{u}^{2}+2huv+b{v}^{2}+2gu+2fv+c$

In this question we have :

$u=3,v=-1,a=2,h=0,b=3,g=\frac{1}{2},h=-\frac{1}{2},c=-5$

Writing the equation we have :

${(\frac{13x-7y-6}{2})}^{2}=20.(2{x}^{2}+3{y}^{2}+x-y-5)$

Simplifying it we get :

$9{x}^{2}-191{y}^{2}-236x+164-182xy+436=0$

Factorising it we get :

$-(9x-191y-218)(x+y-2)=0$

We can simply say that the lines are :

$x+y-2=0$

$9x-191y-218$

So we get :

$a=b=1,c=2,d=9,e=191,f=218$

Finally we get :

$\boxed{\frac{d+e+f-2}{a+b+c}=104}$

$Let\quad the\quad equation(s)\quad of\quad the\quad line(s)\quad tangent\quad to\quad \\ \\ 2{ x }^{ 2 }+3{ y }^{ 2 }+x-y-5=0\quad be\quad y=mx+b\\ \\ Since\quad y=mx+b\quad passes\quad through\quad P(3,-1),\\ \\ \quad \quad \quad y=m(x-3)-1\\ \\ Let\quad A=x-3\\ \\ Substitute\quad y=mA-1\quad to\quad 2{ x }^{ 2 }+3{ y }^{ 2 }+x-y-5=0,\\ \\ 2{ x }^{ 2 }+3{ (mA-1) }^{ 2 }+x-mA+1-5=0\\ 2{ x }^{ 2 }+3({ m }^{ 2 }{ A }^{ 2 }-2mA+1)+x-mA-4=0\\ 2{ x }^{ 2 }+3{ m }^{ 2 }{ A }^{ 2 }-7mA+x-1=0\\ 2{ x }^{ 2 }+3{ m }^{ 2 }({ x }^{ 2 }-6x+9)-7m(x-3)+x-1=0\\ 2{ x }^{ 2 }+3{ m }^{ 2 }{ x }^{ 2 }-18{ m }^{ 2 }x+27{ m }^{ 2 }-7mx+21m+x-1=0\\ { x }^{ 2 }(2+3{ m }^{ 2 })+x(-18{ m }^{ 2 }-7m+1)+27{ m }^{ 2 }+21m-1=0\\ \\ Since\quad the\quad line\quad is\quad tangent\quad the\quad discriminant,\quad D,\quad \\ is\quad equal\quad to\quad 0\\ \\ D={ (-18{ m }^{ 2 }-7m+1) }^{ 2 }-4(2+3{ m }^{ 2 })(27{ m }^{ 2 }+21m-1)\\ =(324{ m }^{ 4 }+252{ m }^{ 3 }+13{ m }^{ 2 }-14m+1)-4(81{ m }^{ 4 }+63{ m }^{ 3 }+51{ m }^{ 2 }+42m-2)\\ =13{ m }^{ 2 }-204{ m }^{ 2 }-14m-168m+1+8\\ =-191{ m }^{ 2 }-182m+9=0\\ \\ -191{ m }^{ 2 }-182m+9=0\\ 191{ m }^{ 2 }+182m-9=0\\ \\ m=\frac { -91\pm \sqrt { 8281+1719 } }{ 191 } \\ \\ m=\frac { -91+100 }{ 191 } ,\quad \frac { -91-100 }{ 191 } \\ \\ m=\frac { 9 }{ 191 } ,\quad -1\\ \\ Since\quad y=m(x-3)-1,\\ Substituting\quad m=\frac { 9 }{ 191 } ,\\ y=\frac { 9x }{ 191 } -\frac { 27 }{ 191 } -1\\ 191y=9x-27-191\\ -9x+191y=-218\\ 9x-191y-218=0\\ The\quad equation\quad 9x-191y-218=0\quad is\quad in\quad the\quad form\quad \\ dx-ey-f=0\\ where\quad d>0,\quad e>0,\quad f>0,\quad and\quad gcd(d,e,f)=1\\ d=9,\quad e=191,\quad f=218\\ \\ Substituting\quad m=-1,\\ y=-x+3-1\\ y=-x+2\\ x+y-2=0\\ The\quad equation\quad x+y-2=0\quad is\quad in\quad the\quad form\quad ax+by-c=0\\ where\quad a>0,\quad b>0,\quad c>0,\quad and\quad gcd(a,b,c)=1\\ a=1,\quad b=1,\quad c=2\\ \begin{cases} a=1 \\ b=1 \\ c=2 \\ d=9 \\ e=191 \\ f=218 \end{cases}\\ \\ \frac { d+e+f-2 }{ a+b+c } =\quad \frac { 9+191+218-2 }{ 4 } =\frac { 416 }{ 4 } =104\\ \\ Therefore,\\ \boxed { \frac { d+e+f-2 }{ a+b+c } =104 } \\ \\ \\ \\ \\$

Considere $r : ax + by - c = 0$ e $s: dx - ey - f = 0$ ,

Da figura conclui-se que $(2, 0) \in r$ , segue que:

$\begin{vmatrix} x & y & 1 \\ 2 & 0 & 1 \\ 3 & - 1 & 1 \end{vmatrix} = 0 \\\\ - x + 2y + 2 - 3y = 0 \\ \boxed{- x - y + 2 = 0}$

Ora, $\boxed{a = 1}$ , $\boxed{b = 1}$ e $\boxed{c = 2}$ .

Dado o ponto $(3, - 1) \in s$ , temos que $y = m(x - 3) - 1$ é a equação da tangente à elipse.

Substituindo a equação da reta $s$ na equação da elipse, devemos encontrar apenas um ponto (tangente) de intersecção, ou seja, $\Delta = 0$ . Segue que,

$2x^2 + 3\left [ m(x - 3) - 1 \right ]^2 + x - \left [ m(x - 3) - 1 \right ] - 5 = 0 \\ (3m^2 + 2)x^2 + (1 - 7m - 18m^2)x + (27m^2 + 21m - 1) = 0$

Anulando o discriminante,

$191m^2 + 182m - 9 = 0 \\ (m + 1)(m - \frac{9}{191}) = 0$

Se fizeres $m = - 1$ , então $e < 0$ ; portanto, $\boxed{m = \frac{9}{191}}$

Segue que,

$y = \frac{9}{191} \cdot (x - 3) - 1 \\\\ 9x - 191y - 218 = 0 \\\\ \begin{cases} d = 9 \\ e = 191 \\ f = 218 \end{cases}$

Com efeito,

$\frac{d + e + f - 2}{a + b + c} = \\\\ \frac{416}{4} = \\\\ \boxed{\boxed{104}}$