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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..

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With the usual informality in these matters: it looks fine to me. – DonAntonio May 14 '13 at 16:05
With a little less fuss, you can also say that since a line perpendicular to the x-axis is "vertical", its slope is "undefined". You can then just look for values of the parameter that make the denominator of the rational function for $\frac{dy}{dx}$ here equal to zero. – RecklessReckoner May 14 '13 at 17:48

1 Answer 1

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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 far i could not find)....thanking you that you yourself solved it...:)

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