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TANGENTS AND NORMALS show that the following curve intersect orthogonally at the indicated points y^2=8x and 2x^2 + y^2 = 10 at (1, 2sqrt(2))

TANGENTS AND NORMALS
 
show that the following curve intersect orthogonally at the indicated points
y^2=8x and 2x^2 + y^2 = 10  at (1, 2sqrt(2))

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello student,
Please find the answer to your question below


y2= 8x
Slope of tangent:
2yy’ = 8
y’ = 4/y............(1)
2x2+ y2= 10
Slope of tangent:
4x + 2yy’ = 0
y’ = -2x/y..........(2)
(1).(2) = -8x/y2
(1).(2) = -8/8 = -1
Hence proved.

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