For the curve x = t$^2 – 1, y = t^2 – t$, the tangent line is perpendicular to x-axis, where
Options are :
a ) t = 0
b) $t \to \infty$
c) $t = \frac{1}{\sqrt{3}}$
d) $t = \frac{-1}{\sqrt{3}}$
Here we have $\frac{dx}{dt} = 2t$ $\& $ $ \frac{dy}{dt} = 2t-1$
$\therefore, \frac{dy}{dx} = \frac{2t-1}{2t}$ If this tangent is parallel to x axis then $\frac{dy}{dx}=0$ and if this is perpendicular then $\frac{-dx}{dy} =0$ $\Rightarrow \frac{2t}{2t-1}=0 \Rightarrow 2t = 0 \Rightarrow t =0$
I hope this is the correct approach please suggest.. thanks..
Best Answer
Yes,you are absolutely right in your procedure to find first dy/dx and then,as it denotes the slope of the tangent at a point,equating it to -(infinity) is the right way....however,let's see if anyone point out any mistake(if any remains...so far i could not find)....thanking you that you yourself solved it...:)