Grade 11Algebra

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

Edward

6 Years agoGrade 11

2 Answers

Arun

6 Years 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

6 Years 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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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

Edward

2 Answers

Arun

Aditya Gupta

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note -4a2 is negative four a squared
y2 is y squared