P(2at , at^2) Q(2au, au^2)
y = tx - at^2
y = ux - au^2
point of intersection R (a(u + t) , aut)
draw a diagram.
plot R on your plane.
and draw two tangents to the x^2 = 4ay
if you were to derive and find the general gradient of all tangents to the parabola
you'd get
dy/dx = x/2a
from your diagram you'd notice that the gradients are same but one gradient has to + and the other has to -.
the general equation of tangents was already given to you in part i) and the point of intersection (general point) was given again.
and from part ii) they say that the two tangents intersect at (a , -6a)
so just equate this to the general point of intersection
a( u + t ) = a
aut = -6a
a simple case of substitution and you will get values for u and t, end up being
u = 3 or -2
t = -2 or 3
sub those values into
P( 2at , at^2) Q ( 2au , au^2)
and then from your diagram draw chord PQ, and then you will see a triangle appear.
and from there just use the distance formula to prove that either of the sides of the triangle are equal and voila, bob is your uncle, you proven the triangle is an isosceles.
(y)
PS : have you noticed that at the end of every post, i leave a (y) which is the same as my avatar (thumbs up).
gosh, i am so witty
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