Eliminate 0 from the following equations.
(i) x= a sec 0, y=b tan0
(ii) x=k + a cos0 , y= h + b sin 0
Answers
Step-by-step explanation:
Given:-
X = a sec 0 and Y = b tan 0
To find:-
Eliminate 0 from the above equations?
Solution:-
X = a sec 0 -----------------(1)
On squaring both sides then
X^2 = a^2 Sec^2 0
X^2/a^2 = Sec^2 0--------(2)
Y = b tan 0 ---------------- -(3)
On squaring both sides
=>Y^2 = b^2 Tan^2 0
=>Y^2/b^2 = Tan^2 0 ----(4)
On Subtracting (4) from (2) then
(X^2/a^2) - (Y^2/b^2) = Sec^2 0 - Tan^2 0
We know that
Sec^2 A - Tan^2 A = 1
=>(x^2/a^2)-(y^2/b^2) = 1 (or)
=>(x^2b^2-y^2a^2)/(a^2b^2) = 1
x^2b^2-y^2a^2 = a^2b^2
Answer:-
x^2b^2-y^2a^2 = a^2b^2
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ii)
Given:-
x=k + a cos0 , y= h + b sin 0
To find:-
Eliminate 0 from the above equations?
Solution:-
x=k + a cos 0------------(1)
=>x-k = a Cos 0
=>Cos 0 = (x-k)/a
On squaring both sides then
=>Cos^2 0 = [(x-k)/a]^2 -----(2)
y= h + b sin 0 ------------(3)
=>y-h = b sin 0
=>Sin 0 = (y-h)/b
On squaring both sides then
=>Sin^2 0 = [(y-h)/b]---------(4)
On adding (4)&(3) then
=>Sin^2 0 + Cos^2 0 = [(x-k)/a]^2 + [(y-h)/b]^2
We know that
Sin^2 A + Cos^2 A = 1
=>1 =[ (x-k)^2/a^2]+[(y-h)^2/b^2]
=>1 = [(x-h)^2b^2+(y-h)^2a^2]/a^2b^2
=>[(x-h)^2b^2+(y-h)^2a^2] = a^2b^2
Answer:-
[(x-h)^2b^2+(y-h)^2a^2] = a^2b^2
Used formulae:-
- Sec^2 A - Tan^2 A = 1
- Sin^2 A + Cos^2 A = 1