Find the equation of the straight line which is both a tangent and normal to the curve $x=3t^2,y=2t^3?$
I found $\frac{dy}{dx}=t$.Let the curve has tangent at point $P(t_1)i.e.(3t_1^2,2t_1^3)$ and let the curve has same line as normal at point $Q(t_2)i.e.(3t_2^2,2t_2^3)$
$t_1=\frac{-1}{t_2}$
But i could not solve further.Please help me.
When you computed $ \frac{dy}{dx} = t, $ you crossed out $t$ from numerator and denominator under the assumption that $ t \ne 0. $ and you are dealing exactly with that situation!
This is a semi-cubical "parabola" which should be sketched to understand tangent behavior at the origin. The tangent rotates to have slopes $ 0, - \infty, 0 $ at the cusp point where a fast rotation and slope transition occur. The tangent appears to be tangent and normal at the narrow cusp.
It is with this view the question makes sense.
You've only related the slopes of the line, but not used the fact that the line is determined by the points it is passing through. The equation of the line will be : $$\frac{y-2t_1^3}{x-3t_1^2}=\frac{2t_1^3-2t_2^3}{3t_1^2-3t_2^2}$$ $$y-2t_1^3=(x-3t_1^2)\left(\frac 23\cdot\frac{t_1^2+t_2^2-1}{t_1+t_2}\right)$$ Now equate the slope of this line to $t_1$ or $-\frac 1 {t_2}$ and you will obtain their values.
