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
y−x−c=0;
n(-1,1). Let’s find intersection parabola and line.
y2+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∨y2+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.
x2−2xy+y2−3x+2y+3=0.
Ax2+Bxy+Cy2+Dx+Ey+F=0
eccentricity is equal
2∗(a−c)2+b2/((a+c)2+(a−c)2+b2)
Eccentricity is equal 2^0.5 and it is greater than 1. So it is hyperbola
Comments
Leave a comment