Thread: locus of tangents

RHS show that the tangents at the extremities of any focal chord of a parabola intersect at right angles at directrix.
My work:
Consider ends of focal chord as P( )and Q( ) .
now point of intersection of tangents are( ).now we know hence point of intersection are .clearly this lies on directrix.
but my second method creates a problem .we know that x-co-ordinate of point of intersection of tangents at P and Q on the parabola is the GEOMETRIC MEAN of the x coordinate of P and Q and y coordinate is the ARITHMETIC MEAN of y coordinates(by definition of book) .therefore for point of intersection I take the GM of x coordinate as hence answer must be (x coordinate)as .
Where is the error.is the definition of book is true for GM and AM for some special case.

Originally Posted by ayushdadhwal

RHS show that the tangents at the extremities of any focal chord of a parabola intersect at right angles at directrix.
My work:
Consider ends of focal chord as P( )and Q( ) .
now point of intersection of tangents are( ).now we know hence point of intersection are .clearly this lies on directrix.
but my second method creates a problem .we know that x-co-ordinate of point of intersection of tangents at P and Q on the parabola is the GEOMETRIC MEAN of the x coordinate of P and Q and y coordinate is the ARITHMETIC MEAN of y coordinates(by definition of book) .therefore for point of intersection I take the GM of x coordinate as hence answer must be (x coordinate)as .
Where is the error.is the definition of book is true for GM and AM for some special case.

There isn't any error. You stopped only one step before the finish:

Since the GM becomes

There is a geometry argument without any algebra for this problem.
Denote the two extremities of the focal chord as A and B, let M be the midpoint of the segment AB. Draw a segment MH which is parallel to the x-axis, crossing the directrix at H.
Draw AD and BE parallel to the x-axis, crossing the directrix at D, E respectively. Connect AH and BH. Let F be the focal point.
Then we have AB=AF+FB=AD+BE=2MH. So we have AM=MB=MH. So AHB is a right triangle. We need only to show that AH and BH are tangent lines. This is quite easy using the optical properties of the parabola.

sir here so if one use this then answer is a .
also sir GM is applicable only when numbers are positive .so how can we apply GM here when we are not sure about the coordinate that they are positive or negative .

Originally Posted by ayushdadhwal

sir here so if one use this then answer is a .
also sir GM is applicable only when numbers are positive .so how can we apply GM here when we are not sure about the coordinate that they are positive or negative .

The equation of the parabola is given as:



If then all x-values must be positive or zero because .

If then all x-values must be negative or zero because .

In both cases the x-values have the same sign which is sufficient for the use of the GM.

For two negtive number $x_1,x_2$, we can get $\sqrt{x_1x_2}$ of course.

If some one believes that $\sqrt{x_1x_2}$ is the geometric mean of two negtive number $x_1,x_$, then we will get the conclusion that in some cases the "mean" is greater than every memeber. Then how can we call it the "mean"?

If some one believes that $-\sqrt{x_1x_2}$ is the geometric mean of two negtive number $x_1,x_$, then how can we explain the inequality AM >= GM?