Question

Cartesian form of the curve $\sqrt{r}=\sqrt{a}\cos \frac{\theta }{2},a>0$ is

• A. $4(x^2+y^2)(x^2+y^2+ax)=a^2{y}^2$
• B. $4(x^2+y^2)(x^2+y^2-ax)=a^2{y}^2$
• C. $4(x^2-y^2)(x^2+y^2-ax)=a^2{y}^2$
• D. $4(x^2-y^2)(x^2+y^2+ax)=a^2{y}^2$

Answer 1

Answer: (B)

$4(x^2+y^2)(x^2+y^2-ax)=a^2{y}^2$

Given, $\sqrt{r}=\sqrt{a}\cos \frac{\theta }{2}$
$\Rightarrow 2r=a\left(2{\cos }^2\frac{\theta }{2}\right)$ ....(Squaring both sides)
$\Rightarrow 2r^2=ra(1+\cos \theta )$ ....(Multiplying both sides by $r$)
$\Rightarrow 2r^2=a(r+r\cos \theta )$
$\Rightarrow {\{2r^2=a(r+r\cos \theta )\}}^2=a^2{r}^2$
$\Rightarrow {(2r^2-ax)}^2=a^2{r}^2$
$\Rightarrow 4(r^2{)}^2-4r^{2}ax+a^2{x}^2=a^2(r^2)$
$\Rightarrow 4\left[{(x^2+y^2)}^2\right]-4ax(x^2+y^2)+a^2{x}^2=a^2(x^2+y^2)$
$\Rightarrow 4(x^2+y^2)[x^2+y^2-ax]=a^2{y}^2$
Hence, option B is the correct answer.