Answer to Question #86803 in Analytic Geometry for Sandeep

Question #86803
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.
1
Expert's answer
2019-04-09T08:57:47-0400

A)Firstly we will find a normal vector to the

yxc=0;y-x-c = 0;

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


y2+4y4c=0;y^2 + 4y - 4c = 0;


Line touches parabola, the discriminant is equal 0


D=1616c=0D = 16 - 16c = 0


c=1c = 1

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


y=x1y2+4x=0y = x -1 \lor y^2 +4x = 0

A(-1,-2)

So the equation of line is

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

- vector-equation of line.

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


x22xy+y23x+2y+3=0.x^2-2xy+y^2-3x+2y+3=0.


Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2+Dx + Ey + F = 0

eccentricity is equal

2(ac)2+b2/((a+c)2+(ac)2+b2)\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


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