Hi there MHF,
After an hour of intensive algebra, I am still unable to solve this question.
Could someone please help me with Part A). And also, if they are kind enough, Part B).
2. The equation of the parabola is .
That means the slope of the tangent to the parabola is calculated by
3. Let be a point on the parabola. The tangent at P to the parabola has the equation:
4. The normal in P has the slope . Since the normal passes through P too it has the equation:
5. Now determine the point of intersection of the parabola and the normal.
6. Show that that means the angle .
Having immense troubles here trying to determine the point of intersection of the parabola and the normal.
When you substitute it in, you have on one side, x^2, whilst on the other side, x. It is impossible to make x the subject of the equation.
How do I complete your step 5?
In other attempts, I have tried to use a separate variable for Q, with P being (2ap, ap^2) and Q being (2aq, aq^2). And then using Pythagoras' Theorem to prove true for the right angle. However, the algebra becomes intense and you are unable to eliminate and cancel down to the RHS. Also, I have attempted to use the gradients, then attempting to substitute the given condition of QS = 2PS, however, that fails also with the algebra coming up to powers of 4.
equation of parabola:
equation of normal:
Solve for x:
Apply the quadratic formula:
The radicand is a complete square!
Therefore you'll get:
So Q has the coordinates
To prove the orthogonality of and respectively I would use vectors as I've described in my previous post.