Question

If tangents are drawn to the curve $y=\frac{x}{1+x^2}$ such that the tangents are concurrent at $(1,\frac{1}{2}),$ then

• A. Number of such tangents is $3$.
• B. Each of the tangents has exactly two points in common with the curve.
• C. Abscissa of point of contact for one of the tangents is $1+\sqrt{2}$.
• D. Abscissa of point of contact for one of the tangents is $1-\sqrt{2}$.

Answer 1

Answer: (A), (C), (D)

Abscissa of point of contact for one of the tangents is $1-\sqrt{2}$.

$y=\frac{x}{1+x^2}$
$\frac{dy}{dx}=\frac{1-x^2}{(1+x^2{)}^2}$
Let point of contact be $P(a,b).$
$A(1,\frac{1}{2})$ and $b=\frac{a}{1+a^2}$
Then slope of $PA=$ slope of tangent at $P$
$\Rightarrow \frac{b-\frac{1}{2}}{a-1}=\frac{1-a^2}{(1+a^2{)}^2}$
$\Rightarrow (\frac{a}{1+a^2}-\frac{1}{2})(1+a^2{)}^2=(a-1\left)\right(1-a^2)$
$\Rightarrow (a-1{)}^2(a^2-2a-1)=0$
$\Rightarrow a^2-2a-1=0$ or $(a-1{)}^2=0$
$\Rightarrow a=1\pm \sqrt{2}$ or $a=1$
Thus, there are three possible tangents.