Question

By eliminating $\theta$ from $\sqrt{x}=a{\cos }^2\theta ,\sqrt{y}=a{\sin }^2\theta$ the expression is

• A. $x^2+y^2=a$
• B. $\sqrt{x}+\sqrt{y}=a$
• C. $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
• D. $x^2+y^2=a^2$

Answer 1

The correct option is B $\sqrt{x}+\sqrt{y}=a$

We have,

$\sqrt{x}=a{\cos }^2\theta$ ……. (1)

$\sqrt{y}=a{\sin }^2\theta$ …….. (2)

On adding both equations, we get

$\sqrt{x}+\sqrt{y}=a{\cos }^2\theta +a{\sin }^2\theta$

$\sqrt{x}+\sqrt{y}=a({\cos }^2\theta +{\sin }^2\theta )$

We know that

${\cos }^2\theta +{\sin }^2\theta =1$

Therefore,

$\sqrt{x}+\sqrt{y}=a$

Hence, this is the answer.