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a normal cord of the parabolla y^2=4ax subtending a right angle at the vertex makes an acute angle with x axis,the angle is,,,,,,,,,,,,,,,,,,sir plz explain in detail
Dear jauneet
let end point of the chord is (at12 ,2at1) and (at22 ,2at2)
since it is normal to parabola so t2 = -t1 - 2/t1
equation of normal of parabola y =-tx + 2at + at3
slope is -t1 which is given acute so -t1 >0
or t1 <0
now chord make right angel at origin so {(2at1 -0)/ (at12 -0)}{(2at1 -0)/ (at12 -0)} =-1
t1t2 =-4
t1(-t1 - 2/t1) =-4
t1 =-√2
so angel is =tan-1(√2)
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get youthe answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation.All the best.Regards,Askiitians ExpertsBadiuddin
et end point of the chord is (at12 ,2at1) and (at22 ,2at2)
or t1
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get youthe answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation.All the best.Regards,Askiitians Experts
nd point of the chord is (at12 ,2at1) and (at22 ,2at2)
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