The Geometry of the Parabola (1 Viewer)

[RivalryofTroll]

P(2ap,ap^2) is any point on the parabola x^2 = 4ay other than its vertex. The normal at P meet the parabola again at Q.

Show that the x-coordinate of Q is one of the roots of the quadratic equation: . Then find the co-ordinates of Q.

Do we use the quadratic formula to find the x-value of Q or is there another way?

 

[Carrotsticks]

Sum and Product of roots.

 

[RivalryofTroll]

Sum and Product of roots.

Woops, I meant to find the co-ordinate Q.

 

[Carrotsticks]

To obtain the quadratic, solve the normal simultaneously with the equation of the parabola.

Then use Sum of Roots, because we know ONE of the roots is P, which has x coordinate 2ap.

So if we let the x coordinate of Q be X, then we have:

X + 2ap = -4a/p (using sum of roots formula)

So X = -4a/p - 2ap = Simplify to get whatever.

 

[Sy123]

We know that the normal at P will have two solutions one solution will be P and the other will be Q.

Now finding the equation of the normal at P, I assume you know how to do this, simply differentiate subsitute x=2ap, find negative reciprocal of it to find gradient of normal, using point gradient formula the normal at P should be:



Now to equate the parabola and the line we need to sub in respective y values, the reason why its y and not x will become clear soon:



We have found the equation that has been presented, now we know that the two intersections of the normal to the parabola and the parabola is P and Q, hence since we arrived at that equation, we can show that Q is one of the solutions

Did I explain that properly?

 

[RivalryofTroll]

We know that the normal at P will have two solutions one solution will be P and the other will be Q.

Now finding the equation of the normal at P, I assume you know how to do this, simply differentiate subsitute x=2ap, find negative reciprocal of it to find gradient of normal, using point gradient formula the normal at P should be:



Now to equate the parabola and the line we need to sub in respective y values, the reason why its y and not x will become clear soon:



We have found the equation that has been presented, now we know that the two intersections of the normal to the parabola and the parabola is P and Q, hence since we arrived at that equation, we can show that Q is one of the solutions

Did I explain that properly?

Yeah I got that already but thanks anyways.

To obtain the quadratic, solve the normal simultaneously with the equation of the parabola.

Then use Sum of Roots, because we know ONE of the roots is P, which has x coordinate 2ap.

So if we let the x coordinate of Q be X, then we have:

X + 2ap = -4a/p (using sum of roots formula)

So X = -4a/p - 2ap = Simplify to get whatever.

Thanks guise.

dam dis isnt easy

 

[Sy123]

Would carrot's sum of roots be necessary though? Equating the normal and the parabola will give us P and Q no matter what (because vertex is not counted), so arriving at the equation given would mean that Q has to be one of the roots?