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Given curve is y2 - 16x – 8y = 0 ….… (1)
Let P = (14, 7)
Then equation (1) can be written as y2 – 8y = 16x
or y2 – 8y + 16 = 16x + 16
Or (y – 4)2 = 16(x + 1) ….….…. (2)
This is of the form (y –l)2 = 4a(x-s), where a = 4, s = -1, l = 4.
Let (-1+4t2, 4+8t) be any point on the curve (2).
Then from (2) , we have 2(y-4) dy/dx = 16.
Hence, dy/dx = 8/(y-4)
at (4t2 – 1, 4 + 8t), dy/dx = 1/t
Hence, the equation of normal at (-1 + 4t2, 4+8t) is
y - 4 – 8t = – t(x + 1 – 4t2)
or tx + y – 4 – 8t + t – 4t3 = 0 ….….. (3)
If line passes through the point P(14, 7) then
14t + 7 – 4t3 – 7t – 4 = 0
or 4t3 – 7t -3 = 0
(t + 1)(4t2 – 4t – 3) = 0
Hence, t = -1, (4 ± 8)/8 = -1, 3/2 , -1/2.
when t = -1, foot of the normal is (3, -4)
when t = 3/2, foot of the normal is (8, 16)
when t = -1/2, foot of the normal is (0, 0).
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