Simulataneous Eq, Quad identies (1 Viewer)

[mtsmahia]

Can someone please help with this Q

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

thanks

And also ,

Find the equation of the parabola which passes through the points (-1,0) and (0,-1) and is symmetrical about the Y-axis

 

[untouchablecuz]

Can someone please help with this Q

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

thanks

tangent of the form y=mx+b

solve simultaneously with equation of parabola,

mx+b=x^2-2x+5

x^2-x(m+2)+(5-b)=0

Δ=(m+2)^2-4(1)(5-b)=m^2+4m+4-20+4b=m^2+4m-16+4b

the lines are tangent and thus only touch the parabola; this means that there is a double zero and Δ=0

i.e. m^2+4m-16+4b=0 [1]

the point (1,3) satisfies the equation of the tangent y=mx+b since it passes through it,

i.e. 3=m+b => b= 3-m [2]

sub [2] in [1]

m^2+4m-16+12-4m=0

m^2-4=(m-2)(m+2)=0

m = +-2

from [2],

when m=2, b=1 => tangent y=2x+1

when m=-2, b=5 => tangent y=-2x+5

 

[cutemouse]

Find the equation of the parabola which passes through the points (-1,0) and (0,-1) and is symmetrical about the Y-axis

Let the parabola be of form: x² = 4a(y-k)

Through (-1,0): 1 = 4a(-k) So 4ak = -1 --(1)

Through (0,-1): 0 = 4a(-1-k) -- (2)

Sub (1) into (2): 0 = 4a(4ak - k)

So 4ak(4a-1)=0 So a=1/4

Therefore x² = y-k

Through (-1,0): 1 = 0 -k, so k=-1

So equation of the parabola is: x²=y+1.



Alternative method:

As (-1,0) lies on this parabola, it cuts the x axis at x=-1. As the parabola is symmetrical about the y-axis it also cuts the x-axis at x=1

So a possible equation of the parabola is: y=a(x-1)(x+1)

Through (0,-1): -1 = a(-1)(1) So a=1

So equation of the parabola is: y=(x-1)(x+1), ie. y=x²-1 ie. x²=y+1.