Question

x = h + a cos t ,y = k + sin t. Eliminate t

Answer 1

Answer :

Given that,

x = h + a cost

⇒ cost = (x - h)/a ...(i)

y = k + sint

⇒ sint = y - k

Now, squaring and adding (i) and (ii), we get

cos²t + sin²t = {(x - h)/a}² + (y - k)²

⇒ 1 = {(x - h)/a}² + (y - k)²

⇒ {(x - h)/a}² + (y - k)² = 1,

which is the required relation after elimination of t.

#MarkAsBrainliest

Answer 2

Hi there!

Given :-

♦ x = h + a cos t

⇒ cos t = $\frac{x \: - \: h}{a}$ ----(i)

♦ y = k + sin t

⇒ sin t = y - k. ---(ii)

Adding n' Squaring eqn. (i) + (ii) :-

sin² t + cos² t = $\frac{x \: - \: h}{a}$ + (y - k)²

⇒ 1 = $\frac{x \: - \: h}{a}$ + (y -k)²

⇒ $\frac{x \: - \: h}{a}$ + (y -k)² = 1

[ Required solution after elimination of t. ]

hope it helps! :)