Question

1.Find the equation of the normal to the parabola y^2+4x=0 at the point where the line y=x+c touches it. Q.(2) Identify and trace the conic x^2-2xy+y^2-3x+2y+3=0.

Answer 1

A)Firstly we will find a normal vector to the

$$y-x-c = 0;$$

n(-1,1). Let’s find intersection parabola and line.

$$y^2 + 4y - 4c = 0;$$

Line touches parabola, the discriminant is equal 0

$$D = 16 - 16c = 0$$

$$c = 1$$

Let A(x,y) - intersection of line and parabola.

$$y = x -1 \lor y^2 +4x = 0$$

A(-1,-2)

So the equation of line is

$$X = A+k*n;$$

- vector-equation of line.

B)The maximum power of equation is 2. So it is conic anyway. 

$$x^2-2xy+y^2-3x+2y+3=0.$$

$$Ax^2 + Bxy + Cy^2+Dx + Ey + F = 0$$

eccentricity is equal

$$\sqrt{2*\sqrt{(a-c)^2+b^2}/((a+c)^2+\sqrt{(a-c)^2+b^2})}$$

Eccentricity is equal 2^0.5 and it is greater than 1. So it is hyperbola