Vikas TU
Last Activity: 7 Years ago
This condition to the hyperbola is given as
x2/a2 – y2/b2 = 1 … (1)
Let P any point on it as (a sec θ, b tan θ), at that point the condition of digression at P is
x/a – y/b sin θ = cos θ … (2)
The condition to the asymptotes to (1) are
x/a = y/b … (3)
what's more, x/a = – y/b … (4)
Explaining (2) and (3), we get the directions of Q as
(a cos θ/1–sin θ, bcosθ/1–sinθ)
Understanding (2) and (4), we get the directions of R as
(acosθ/1+sinθ, –bcosθ/1+sinθ)
Give O a chance to be the focal point of the hover going through C, Q and R having its directions as (h, k). At that point obviously OC = OQ
⇒ h2 + k2 = (h–acosθ/1–sinθ)2 + (k–bcosθ/1–sinθ)2
⇒ h2 + k2 = (h–acosθ/1–sinθ)2 + (k–bcosθ/1–sinθ)2
⇒ h2 + k2 = h2 + k2 + (a2 + b2) cos2θ/(1–sinθ)2 – (2ah + 2bk) cosθ/1–sinθ
⇒ 2(ah + bk) = (a2 + b2) cosθ/1–sinθ … (5)
So also OC = OR
Consequently h2 + k2 = (h–acosθ/1+sinθ)2 + (k+bcosθ/1+sinθ)2
Which on rearrangements as in the last case, given 2(ah –bk) = (a2 + b2) cosθ/1–sinθ … (6)
to get the locus of the direct O we have toward take out f from (5) and (6), so increasing the two we get
4(a2h2 – b2k2) = (a2 + b2) cos2θ/1–sin2θ = (a2 + b2)
for (h, k), we get the required locus as
4(a2x2 – b2y2) = (a2 + b2)2
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