If x = h + a cos θ, y = k + b sin θ, then prove that : ((x−h)/a)2

Q) If x = h + a cos θ, y = k + b sin θ, then prove that : $(\frac{x - h}{a})^2+ (\frac{y - k}{b})^2 = 1$

(Q 26 A – 30/1/3 – CBSE 2026 Question Paper)

Ans:

Step 1: We are given: x = h + a cos θ

∴ x – h = a cos θ …………… (i)

Also, we are given, y = k + b sin θ

∴ y – k = b sin θ ……………..(ii)

Step 2: Let’s start from LHS:

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

Substituting values from eqaution (i) and (ii):

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

= $(\frac{a \cos \theta}{a})^2+ (\frac{b \sin \theta}{b})^2$

= $(\frac{\cancel{a} \cos \theta}{\cancel{a}})^2+ (\frac{\cancel{b} \sin \theta}{\cancel{b}})^2$

= cos² θ + sin ² θ

Step 2: Since cos² θ + sin ² θ = 1 (by trigonometric identity)

∴ LHS = 1 = RHS

Hence Proved!

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