Question

If $x^2+y^2=1$, then (IIt-JEE, 2000)

• A. $y{y}^{{}^{\prime \prime }}+(y^{{}^{\prime }}{)}^2+1=0$
• B. $y{y}^{{}^{\prime \prime }}+2(y^{{}^{\prime }}{)}^2+1=0$
• C. $y{y}^{{}^{\prime \prime }}-2(y^{{}^{\prime }}{)}^2+1=0$
• D. $y{y}^{{}^{\prime \prime }}+(y^{{}^{\prime }}{)}^2-1=0$
• E. $y{y}^{{}^{\prime \prime }}-(2y^{{}^{\prime }}{)}^2-1=0$

Answer 1

The correct option is A $y{y}^{{}^{\prime \prime }}+(y^{{}^{\prime }}{)}^2+1=0$

$x^2+y^2=1$

$\Rightarrow 2x+2y{y}^{{}^{\prime }}=0$

$\Rightarrow x+y{y}^{{}^{\prime }}=0$

$\Rightarrow 1+y{y}^{{}^{\prime \prime }}+\left(y^{{}^{\prime }}\right)=0$