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The cordinates of the ends of a focal chord of a parabola y2=4ax are (x1,y1) and (x2,y2). show that x1x2=a2 and y1y2=-4a2 note -4a2 is negative four a squared y2 is y squared

The cordinates of the ends of a focal chord of a parabola y2=4ax are (x1,y1) and (x2,y2). show that x1x2=a2 and y1y2=-4a2
 
note -4a2 is negative four a squared 
y2 is y squared

Grade:11

2 Answers

Arun
25763 Points
one year ago
Dear student
 
Please attach an image of question as it is not clear where is square(power 2) or only 2(multiplied)
Aditya Gupta
2075 Points
one year ago
note that focus of parabola y^2=4ax is (a, 0), and focal chord has eqn
y= m(x – a). solve these eqns simultaneously to find intersection.
so m^2(x^2+a^2 – 2ax)= 4ax
or m^2*x^2 – 2ax(m^2+2) + a^2*m^2= 0
its roots obviously are x1, x2 whose product would be
x1.x2= a^2*m^2/m^2
or x1x2 = a^2
similarly, we could eliminate x and write 
y^2= 4a(y/m + a)
or y^2 – (4a/m)y – 4a^2 = 0, the roots of which obviously are y1 and y2.
so y1*y2 = – 4a^2/1
or y1y2 = – 4a^2
kindly approve :))

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