Question

The equation of curve passing through (1,1) in which the subtangent is always bisected at the origin is

• A. $\frac{1}{\sqrt{e}}$
• B. ${2x}^2-y=1$
• C. $x^2-y=0$
• D. x+y=2

Answer 1

Answer: (C)

$x^2-y=0$

Equation of tangent at (x, y) is

$Y-y=\frac{dy}{dx}(X-x)$

In given figure Tm is known as sub tangent coordinate of m is $(x,0)$.

T is the point where tangent crosses x-axis, hence y will be 0.

$\Rightarrow 0-y=\frac{dy}{dx}(X-x)$

$-\frac{dx}{dy}y=X-x$

$X=-\frac{dx}{dy}y+x$

Hence coordinates of T are $(-\frac{dx}{dy}y+x,0)$

As mT is bisceted at origin, hence origin will be midpoint of mT.

$\Rightarrow \frac{x+(-\frac{dx}{dy}y+x)}{2}=0$

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

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

$\frac{dy}{y}=2\frac{dx}{x}$

$\Rightarrow \ln \left(y\right)+\ln \left(C_1\right)=2\ln \left(x\right)$, where $\ln \left(C_1\right)$ is integration control

$\Rightarrow 2\ln \left(x\right)-\ln \left(y\right)=\ln \left(c_1\right)$

$\Rightarrow \ln \left(x^2\right)-\ln \left(y\right)=\ln \left(c_1\right)$

$\Rightarrow \ln \left(\frac{x^2}{y}\right)=\ln \left(c_1\right)$

$\frac{x^2}{y}=c_1$

As $(1,1)$ lies on curve

$\frac{1^2}{1}=c_1\Rightarrow c_1=1\Rightarrow \frac{x^2}{y}=1$

$y=x^2$