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Grade: 11

                        

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

9 months ago

Answers : (2)

Arun
25359 Points
							
Dear student
 
Please attach an image of question as it is not clear where is square(power 2) or only 2(multiplied)
9 months ago
Aditya Gupta
2069 Points
							
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 :))
9 months ago
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