Question

Let $y=f\left(x\right)$ be a vertex whose parametric equation is $x=(t^2+t+1),y=(t^2-t+1)$ where $t>0$. The total number of tangents that can be drawn to this curve from $(1,1)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $Noneofthese$

Answer 1

The correct option is A $1$

Consider $\frac{dy}{dt}=2t-1,\frac{dx}{dt}=2t+1⟹\frac{dy}{dx}=\frac{2t-1}{2t+1}$

Equation of tangent has slope $\frac{2t-1}{2t+1}$ and passing through $(1,1)$

$y-1=\frac{2t-1}{2t+1}(x-1)$

This equation passes through $(t^2+t+1,t^2-2t+1)$

$⟹(t^2-2t)(2t-1)=(2t+1)(t^2+t)$

$⟹t=0,1$

Since $t>0$

There is only one possible tangent that can be drawn