Question

The equation of curve passing through (1, 1) in which the sub-tangent is always bisected at the origin cannot be

• A. $y^2=x$
• B. $2x^2-y^2=1$
• C. $x^2+y^2=2$
• D. $x+y=2$

Answer 1

Answer: (B), (C), (D)

The correct options are
B

$2x^2-y^2=1$

C

$x^2+y^2=2$

D

$x+y=2$

Let TM be the sub-tangent where $T\equiv (x-y\frac{dx}{dy},0)$ and $M\equiv (x,0)$

$\because$ sub tangent is bisected at origin $\Rightarrow \frac{1}{2}(x-y\frac{dx}{dy}+x)=0$

$\Rightarrow 2x-y\frac{dx}{dy}=0\Rightarrow 2\int \frac{dy}{y}=\int \frac{dx}{x}$

$\Rightarrow 2\ln =\ln cx$

$y^2=cx$

curve passes through $(1,1)\therefore c=1,y^2=x.$