The expression is also dividable by q-p.sooyong94 said: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
Homework Equations
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
There should be minus in front of the last term in the first equation. You made a mistake when copying.sooyong94 said:Strangely enough I got this:
You have to work with x and y. What are p and q now?sooyong94 said:
Have you copied the question correctly? These x and y values do not fulfill the equation given y^2(x+2a)+4a^3 =0.sooyong94 said:
I mean the problem might be wrong. The coordinates of the point of intersection do not fit to the given locus.sooyong94 said:
A tangent line on a parabola is a straight line that touches the curve of the parabola at only one point. This point of contact is called the point of tangency.
The slope of the tangent line is equal to the slope of the parabola at the point of tangency. This is because the tangent line and the parabola share the same direction at that point.
The equation of a tangent line on a parabola can be written in the form y = mx + b, where m is the slope of the tangent line and b is the y-intercept. To find the slope, you can use the derivative of the parabola's equation.
A normal line on a parabola is a straight line that is perpendicular to the tangent line at the point of tangency. It intersects the parabola at the point of tangency and forms a right angle with the tangent line.
The slope of the normal line can be found by taking the negative reciprocal of the slope of the tangent line. Then, the equation can be written in the form y = mx + b, where m is the slope of the normal line and b is the y-intercept, which can be found by plugging in the point of tangency into the equation.