Solving for t When the Normal Chord at a Point on the Parabola Subtends a Right Angle at the Vertex

This article delves into the mathematical concept of the normal chord on a parabola and the conditions under which it subtends a right angle at the vertex. We'll explore the parametric form of the parabola equation, derive the equation of the normal, and find the value of ( t ) using a systematic approach. The article also includes alternative methods to confirm the solution.

Introduction

A parabola is a conic section defined by the quadratic equation $y^2 4ax$. This equation describes the path of a projectile under gravity or the locus of points equidistant from a fixed point (the focus) and a fixed line (the directrix). One interesting property of parabolas is the behavior of normal chords, particularly when they subtend a right angle at the vertex.

Mathematical Setup

To solve for ( t ) in the given problem, we need to understand the geometry and algebra of parabolas. A point on the parabola can be represented in parametric form as:

$P_t (at^2, 2at)$

Determining the Equation of the Normal

The slope of the tangent at the point ( P_t ) can be found by differentiating the equation of the parabola:

$frac{dy}{dx} frac{2a}{2at} frac{1}{t}$

Therefore, the slope of the normal line, which is the negative reciprocal of the tangent, is:

$-t$

The equation of the normal at point ( P_t ) can be derived using the point-slope form:

$y - 2at -t(x - at^2)$

Rearranging, we get:

$y -tx at^3 2at$

Intersection Points

To find where the normal intersects the parabola again, substitute the expression for ( y ) back into the parabola equation:

$(-tx at^3 2at)^2 4ax$

Expanding and simplifying this, we get a quadratic equation in ( x ), which can be solved to find the intersection points.

Right Angle Condition

The normal chord subtends a right angle at the vertex if the product of the slopes of the two lines from the vertex to the endpoints of the chord equals (-1). This condition can be applied to the slopes calculated from the intersection points.

Solving for t

The conditions derived from the slopes lead to a quadratic equation in ( t ). Solving this equation yields the values of ( t ) that satisfy the condition. Through the analysis and solving, we find that the values of ( t ) are:

$t 1$ and ( t -1$

Note: This result can also be obtained by another method: solving the equations ( y^2 4ax ) and ( y x ), or ( y -x ).

Alternative Method

To confirm the solution, consider the following alternative method:

Solve $y^2 4ax$ and $y x$. Substituting into the parabola equation, we get:

$x^2 4ax$

This gives:

$x(x - 4a) 0$

So, ( x 0 ) or ( x 4a ). The point ( x 0, y 0 ) only gives a single point and doesn't define a subtended angle. Thus, using ( x 4a, y 4a ), and ( x 4a, y -4a ).

Therefore, the values of ( t ) that satisfy the condition are: