# Tangent line and normal on a parabola

## Homework Statement

If the normal at P(ap^2 ,2ap) to the parabola y^2 = 4ax meets the curve again at Q(aq^2, 2aq), show that p^2 +pq+2=0

Point-slope form

## The Attempt at a Solution I tried putting y=2aq and x=aq^2 but I can seem to simplify the whole thing other than dividing both sides by a

• ehild

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ehild
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## Homework Statement

If the normal at P(ap^2 ,2ap) to the parabola y^2 = 4ax meets the curve again at Q(aq^2, 2aq), show that p^2 +pq+2=0

Point-slope form

## The Attempt at a Solution I tried putting y=2aq and x=aq^2 but I can seem to simplify the whole thing other than dividing both sides by a
The expression is also dividable by q-p.

Strangely enough I got this: ehild
Homework Helper
Strangely enough I got this: There should be minus in front of the last term in the first equation. You made a mistake when copying.

Thanks - worked that out quickly.

Now I'm stuck on the second part:

Show that the equation of the locus of the point of intersection of the tangents at P and Q to the parabola is y^2(x+2a)+4a^3 =0. What does this mean? Does this mean that the tangents at P and Q meet at a point?

I managed to find the points of intersection of the two tangents (apq, a(p+q)), but I can't seem to continue at this point.

ehild
Homework Helper
Show that the equation of the locus of the point of intersection of the tangents at P and Q to the parabola is y^2(x+2a)+4a^3 =0. What does this mean? Does this mean that the tangents at P and Q meet at a point?

I managed to find the points of intersection of the two tangents (apq, a(p+q)), but I can't seem to continue at this point.
You have to work with x and y. What are p and q now?

x=apq
y=a(p+q)

ehild
Homework Helper
x=apq
y=a(p+q)
Have you copied the question correctly? These x and y values do not fulfill the equation given y^2(x+2a)+4a^3 =0.

I'm sorry - but I can't catch it.

ehild
Homework Helper
I'm sorry - but I can't catch it.
I mean the problem might be wrong. The coordinates of the point of intersection do not fit to the given locus.