Quadratic Functions (1 Viewer)

[Petinga]

help plz

1. for what values of c is the line y= x+c a tangent to the circle x^2 + y^2=4?
find the coordinates of the point of contact.

2. Find the equation of the two lines which contain the point (1,3) and are tangents to the parabola y= x^2 - 2x +5.

 

[word.]

1. if y = x + c is a tangent to x² + y² = 4,
solve simultaneously:
x² + x² + 2xc + c² = 4
2x² + 2xc + c² - 4 = 0
for a tangent there is only one solution i.e. discriminant {b² - 4ac} = 0
4c² - 8(c² - 4) = 0
-4c² + 32 = 0
c² = 8
c = +-2Sqrt(2)

y = x +- 2Sqrt(2)
x² + y² = 4
x² + x² +- 4xSqrt(2) + 8 = 4
2x² +- 4xSqrt(2) + 4 = 0

x = +- Sqrt(2)
2 + y² = 4
y = +- Sqrt(2)
coords are [-Sqrt(2),Sqrt(2)] and [Sqrt(2),-Sqrt(2)]


2.
y - 3 = m(x - 1)
mx - y + 3 - m = 0
y = mx - m + 3
y = x² - 2x + 5
solve simultaneously for one solution

mx - m + 3 = x² - 2x + 5
x² - x(2 + m) + (2 + m) = 0

(2 + m)² - 4(2 + m) = 0
4 + 4m + m² - 8 - 4m = 0
m² = 4
m = +- 2

y - 3 = -2(x - 1); 2x + y - 5 = 0
y - 3 = 2(x - 1); 2x - y + 1 = 0

 

[Trev]

1)
Sub y=x+c into equ. circle so
x²+(x+c)²=4
2x²+2cx+c²-4=0
There must be two real roots of this equation for intersection so let discriminate equal zero
(2c)²-4.2.(c²-4)=0
-4c²+32=0
When c = +/-2√2 the line is a tangent, tangent is y=x+/-2√2.
then solve 2x²+4√2x+4=0 for x coordinates of intersection, then find y coordinates.