wiz-icon
MyQuestionIcon
MyQuestionIcon
1822
You visited us 1822 times! Enjoying our articles? Unlock Full Access!
Question

Prove that the normal chord at the point whose ordinate is equal to its abscissa subtends a right angle at the focus.

Open in App
Solution


For y2=4ax when ordinate is equal to abscissa
y2=4ayy=4a,0x=4a,0
So the point be P (4a,4a)
Any point on parabola is of the form (at21,2at1)
2at1=4at1=2
If the normal intersect the parabola again then
t2=t12t1
t2=3
Let it intersect again at Q(at22,2at2)
Q(9a,6a)
Focus of the parabola is S(a,0)
Slope of PS=m1=4aa4a0=34
Slope of QS=m2=9aa6a0=43
m1×m2=34×43=1
Hence proved that lines are perpendicular
So the normal chord subtends right angle at the centre

698307_641468_ans_08f21d86e16842f5bb51d880c2e91030.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Jumping to Details
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon