So we are doing parametric equations in class, and there are various fomulas for finding a number of things out about a parabola including the point of intersection of normals, of tangents, the gradients and equations of normals and tangents, the equation of a chord etc...
So I was wondering if there was a formula for the point(s) of intersection between a parabola and a normal from a point on the parabola. It makes sense that there should be, and I can solve it by just simultaneously solving the equations of the normal and parabola, but my varied attempts at generating a fomula have failed.
Here is what my formula ended up looking like, without any simplifications :
px^2 - 4ax - 4a^2p*(2 + p^2) = 0
But it didn't work.
Thanks in advance for any help :)
+26396 So I was wondering if there was a formula for the point(s) of intersection between a parabola and a normal from a point on the parabola
1. Parabola: f(x)=y=ax2+bx+c
2. Normal from a point on the parabola: n(x)=y=−1f′(x)⋅(x−xa)+f(xa)
3. f′(xa)=2axa+bya=f(xa)=ax2a+bxa+c
4. Intersection between a parabola and a normal:
ax2+bx+c=−1f′(x)⋅(x−xa)+ya|⋅f′(xa)ax2⋅f′(xa)+b⋅f′(xa)+c⋅f′(xa)=xa−x+ya⋅f′(xa)…x2⋅[a⋅f′(xa)]+x⋅[1+b⋅f′(xa)]+f′(xa)⋅(c−ya)−xa=0…
x1,2=−[ 1+b⋅f′(xa) ]±√1+[ f′(xa) ]2⋅[ 2+b2−4⋅a⋅(c−ya) ]2⋅a⋅f′(xa)ya=a⋅x2a+b⋅xa+cf′(xa)=2a⋅xa+b
5. Example:
a=12b=0c=0xa=4
Parabola: f(x)=12x2
f(xa)=f(4)=ya=12⋅42=8f′(xa)=f′(4)=2⋅12⋅4+0=4
x1,2=−[ 1+0 ]±√1+[ 4 ]2⋅[ 2+02−4⋅12⋅(0−8) ]2⋅12⋅4x1,2=−1±√1+16⋅[ 2−4⋅12⋅(−8) ]4x1,2=−1±√1+16⋅[ 2+16 ]4x1,2=−1±√1+16⋅[ 18 ]4x1,2=−1±174x1=−1+174x1=xa=4x2=−1−174x2=xb=−4.5y2=yb=12⋅x2byb=12⋅(−4.5)2yb=10.125
+26396 continuation
x1,2=−[ 1+b⋅f′(xa) ]±√1+[ f′(xa) ]2⋅[ 2+b2−4⋅a⋅(c−ya) ]2⋅a⋅f′(xa)ya=a⋅x2a+b⋅xa+cf′(xa)=2a⋅xa+bx1+x2=−[ 1+b⋅f′(xa) ]+√1+[ f′(xa) ]2⋅[ 2+b2−4⋅a⋅(c−ya) ]2⋅a⋅f′(xa)+−[ 1+b⋅f′(xa) ]−√1+[ f′(xa) ]2⋅[ 2+b2−4⋅a⋅(c−ya) ]2⋅a⋅f′(xa)x1+x2=−[ 1+b⋅f′(xa) ]2⋅a⋅f′(xa)+−[ 1+b⋅f′(xa) ]2⋅a⋅f′(xa)x1+x2=−2⋅[ 1+b⋅f′(xa) ]2⋅a⋅f′(xa)x1+x2=−[ 1+b⋅f′(xa) ]a⋅f′(xa)x1+x2=−1a⋅[1f′(xa)+b]|f′(xa)=2axa+bx1+x2=−1a⋅(12axa+b+b)|x1=xax2=xbxa+xb=−1a⋅(12axa+b+b)xb=−xa−1a⋅(12axa+b+b)
conclusion
Parabola y=ax2+bx+c and we have a,b,c and xa⇒xb=−xa−1a⋅(12axa+b+b)⇒yb=ax2b+bxb+c
Example:
Parabola y=12x2a=12b=0c=0xa=4xb=−4−112⋅(12⋅12⋅4+0+0)xb=−4−112⋅(14)xb=−4−24xb=−4−12xb=−4.5yb=12⋅(−4.5)2+0⋅(−4.5)+0yb=10.125
