Skip to content
Question 126 of 129

Q.Which of the following equations is the locus of (acos⁡θ,bsin⁡θ)(a\cos\theta, b\sin\theta)?

(a) x2+y2=a2x^2+y^2=a^2
(b) x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1
(c) y2=4axy^2=4ax
(d) x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2026MCQ· 1mImportance★★★★★
98% · 126/129 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

With x=acos⁡θx=a\cos\theta, y=bsin⁡θy=b\sin\theta, using cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1 eliminates θ\theta to give the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.

Let x=acos⁡θx=a\cos\theta and y=bsin⁡θy=b\sin\theta. Then cos⁡θ=xa\cos\theta=\dfrac{x}{a} and sin⁡θ=yb\sin\theta=\dfrac{y}{b}.

Using the Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1:

(xa)2+(yb)2=1  ⇒  x2a2+y2b2=1\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1 \;\Rightarrow\; \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.