equation of the noraml to x^2=4ay (1 Viewer)

[ozidolroks]

Can someone please help me with these questions ?

a) Prove that the line x-2y-2=0 is a tangent to the parabola x^2=16y.
b) Given the parabola x^2 = 1/2, for what values of m does the line y= m(x+1):

i) have 2 points of intersection ?
ii) one point of intersection?
iii) no points of intersection?

thanks

 

[x3.eddayyeeee]

for the first one solve the two equations as simulatneous equtions to prove they intersect as well as getting the coordinates.
. . .
not sure on the second on though. probably similar to the first one.

 

[ozidolroks]

Ummmm, there is 2 parts to that question ... find the point of contact and prove its a tangent.
I know how to find the point of contact but i don't know how to prove it without finding the point of contact .

I hope this makes sense

 

[kaz1]

Can someone please help me with these questions ?

a) Prove that the line x-2y-2=0 is a tangent to the parabola x^2=16y.
b) Given the parabola x^2 = 1/2, for what values of m does the line y= m(x+1):

i) have 2 points of intersection ?
ii) one point of intersection?
iii) no points of intersection?

thanks

wtf?

You probably mean x²=y/2

i)x²=m(x+1)/2
2x²=mx+m
2x²-mx-m=0
Discriminant>0 for 2 points of intersection
m²-4(2)(-m)>0
m²+8m>0
m(m+8)>0
m>0,m<-8

ii)2x²-mx-m=0
Discriminant=0 for one point of intersection
m(m+8)=0
m=0,-8

iii)2x²-mx-m=0
Discriminant<0 for no point of intersection
m(m+8)<0
-8<m<0

 

[ozidolroks]

Prove that the the line x+ y+ 3 =0 is a tangent to the parabola x^2= 12y. What is the point of contact.

 

[tacogym27101990]

Prove that the the line x+ y+ 3 =0 is a tangent to the parabola x^2= 12y. What is the point of contact.

y=-x-3 (1)
x^2=12y (2)
(1) into (2)

x^2=12(-x-6)
x^2= -12x-36
x^2 + 12x + 36 = 0
discriminant= b^2 - 4ac
= 12^2 - 4*36
=0
therefore tangent as it only has one point of contact


now to find the point of contact lets solve x^2 + 12x +36 = 0
which is a perfect square so:
(x+6)^2=0
so x=-6
using (1) y = -(-6) - 3
y=3
therefore p.o.c is (-6, 3)