Question

1. Consider the conic C in the Euclidean plane described by the following equation: (a) Convert this equation into homogeneous coordinates. b) Express this homogeneous equation using a symmetric matrix (c) Show that C is nonsingular. (d) What type of conic is C? (e) Find all points of intersection between the line 0 and C. Find the equations of the tangents to C at these points. 12)

Answer 1

(a) The homogeneous equation is

$z^2(3(x/z)^2+2(y/z)^2+7(x/z)(y/z)+4(x/z)+5(y/z)+3)=0$

which gives

$3x^2+2y^2+7xy+4xz+5yz+3z^2=0$

(b) Let us say that the matrix is

ce bd e

Then, we have

се =(ax + by + cz cx + ey + fz) bx + dy + ez y

which gives . Hence, the matrix is

a = 3. d= 2. f = 3. b = 3.5, c= 2. e= 2.5

3 3.5 2 3.5 2 2.5 2 2.5 3

(c) Since

3.5 25+2det2 2.5 3 3.5 2 2.53.5det23 3.5 2 +2 det - 3.5 det det 3.5 2 2.53 det 2.5 3 2 2.5 3 ー-0.75-19.25 + 9.5 一一10.5 0

the conic is non-singular.

d) Since

3 3.56.75 det 3.52

is negative, the conic is hyperbola.

e) Letting $\begin{align*}x=0\end{align*}$ in the conic equation, we get

(y +1) (2y +3)0

Thus, the points of intersections are $\begin{align*}(0,-1)\end{align*}$ and .

1.5

From the equation we get (via differentiation)

$6x+4yy'+7xy'+7y+4+5y'=0~\Rightarrow~y'(4y+7x+5)=-(6x+7y+4)$

Thus, the slope is

$y'=-{\frac{(6x+7y+4)}{(4y+7x+5)}}$

At $\begin{align*}(0,-1)\end{align*}$ the slope is $y'=-{\frac{(-7+4)}{(-4+5)}}=3$ ; so the tangent is

${\frac {y+1}{x}}=3$

that is, $y=3x-1$. At
1.5
the slope is $y'=-{\frac{(-10.5+4)}{(-6+5)}}=-6.5$ ; so the tangent is

${\frac {y+1.5}{x}}=-6.5$

that is, $y=-6.5x-1.5$.