Simultaneous Parabola (1 Viewer)

[Lukybear]

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

 

[arjiparji]

tangent to parabola = gradient is negetive reciprical of the eqn y = x^2-2x+5 .

Once you have the negetive reciprical, plug the points + gradient into the : point gradient formula

y-y1= m(x-x1)

this may not be 100% correct so prob. ask someone else too

 

[ninetypercent]

dy/dx = 2x - 2






or

dy/dx = 2x - 2
when x = 0, m = -2





when x = 2, m = 2





I think... this must be wrong

 

[Lukybear]

um, no calculus is involved i think. Cause its under quadratic function. + I dont think (1,3) is actually on the parabola.

BTW: The above is right 90%. But i dont get it...

 

[untouchablecuz]

the tangent is in the form:

y=mx+b

solving simultaneously with y=x^2-2x+5

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

discriminant = 0 since the line is tangent and there is only 1 point of contact

i.e. (m+2)^2-4(5-b)=0 => m^2+4m-16+4b=0 [1]

(1,3) satisfies y=mx+b since it passes through it

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

sub [2] in [1]

m^2-4=0

m= 2 or -2

from [2] b = 1 or 5

hence,

y=2x+1 and y=-2x+5 are the tangents

 

[Lukybear]

very nice. Ive got it. THanks very much.

 

[untouchablecuz]

anytime my man

anytimeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee