Question

please answer this question ​

Answer 1

Here's your solution buddy

Answer 2

(Instead of θ, I use A)

Answer :

Step-by-step explanation:

Given :

$x = h + a cos A$

$y = k + b sin A$

To Prove :

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

Proof :

We know that,

$cos^2A+sin^2A =1$

Hence,

$x = h + a cos A$

⇒ $x - h = a cos A$

⇒ $\frac{x - h}{a} = cosA$

⇒ $(\frac{x - h}{a})^2 = cos^2A$ ...(i)

$y = k + b sin A$

⇒ $y - k= b sin A$

⇒ $\frac{y - k}{b} = sinA$

⇒ $(\frac{y - k}{b})^2 = sin^2A$ ..(ii)

By adding (i) & (ii),

We get,

⇒ $(\frac{x - h}{a})^2 + (\frac{y - k}{b})^2 = cos^2A + sin^2A = 1$

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

Hence proved,.