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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

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9 months ago

```							Dear student Please attach an image of question as it is not clear where is square(power 2) or only 2(multiplied)
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9 months ago
```							note that focus of parabola y^2=4ax is (a, 0), and focal chord has eqny= m(x – a). solve these eqns simultaneously to find intersection.so m^2(x^2+a^2 – 2ax)= 4axor m^2*x^2 – 2ax(m^2+2) + a^2*m^2= 0its roots obviously are x1, x2 whose product would bex1.x2= a^2*m^2/m^2or x1x2 = a^2similarly, 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^2/1or y1y2 = – 4a^2kindly approve :))
```
9 months ago
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