Answer to Question #157488 in Calculus for Ivo

Question #157488
Given that the equation of tangent at the point (at^2 , 2at) on the parabola y^2 = 4ax is x - ty + at^2 = 0
A(at1^2 , 2at1) and B(at2^2 , 2at2) are points on this parabola. The equation of the chord AB is 2x - (t1 + t2)y + 2at1at2 = 0
The tangents A and B meet at the point P. Find the coordinates of P.
The line through P parallel to the axis of the parabola meets the chord AB at M. Find the coordinates of M. Prove that M is the midpoint of AB.
1
Expert's answer
2021-02-02T01:44:47-0500

x=y24a,x=\frac{y^2}{4a},

k=xy=y2a,k=x_y'=\frac{y}{2a},

xati2=2ati2a(y2ati),x-at_i^2=\frac{2at_i}{2a}(y-2at_i),

{xat12=t1(y2at1),xat22=t2(y2at2);\begin{cases} x-at_1^2=t_1(y-2at_1), \\ x-at_2^2=t_2(y-2at_2); \end{cases}

{x0=at1t2,y0=a(t1+t2);\begin{cases} x_0=at_1t_2, \\ y_0=a(t_1+t_2); \end{cases} – coordinates of point P.

Line PM (parallel to the x-axis) has an equation y=a(t1+t2).y=a(t_1+t_2).

Midpoint of AB has coordinates (a2(t12+t22), a(t1+t2)).(\frac a2(t_1^2+t_2^2),~a(t_1+t_2)).

AB equation: 2x(t1+t2)y+2at1t2=0,2x-(t_1+t_2)y+2at_1t_2=0,

with a PM equation we'll have

{x0=a2(t12+t22),y0=a(t1+t2);\begin{cases} x_0=\frac a2(t_1^2+t_2^2), \\ y_0=a(t_1+t_2); \end{cases} – coordinates of point M, and they are equal to the coordinates of point P.


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