Question

Find the equation of the straight line which is tangent at one point and normal at another point of the curve $x=3t^2,y=2t^3$

Answer 1

Solution:-

The slope of the tangent at P(t) is-

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=t$

Let the tangent cuts the curve and normal at $Q\left(t_1\right)$, then the coordinates of the point is $(3{t_1}^2,2{t_1}^3)$

The the slope of the line passing throught the points P $\&$ Q is $\frac{(2{t_1}^3-2t^3)}{(3{t_1}^2–3t^2)}=t$

On solving the above equation, we get

$t_1=-\frac{t}{2}$

$\therefore$ the coordinates of the Q $=(\frac{3t^2}{4},-\frac{t^3}{4})$

Slope of the normal at Q is-

$\frac{dx}{dy}=\frac{2}{t}$

The equation of the straight line having slope t and passing throught Q is-

$y-(-\frac{t^3}{4})=t(x-\frac{3t^2}{4})$

$\Rightarrow tx-y=t^3..........\left(1\right)$

The equation of the straight line having slope $\frac{2}{t}$ and passing throught Q is-

$y-(-\frac{t^3}{4})=\frac{2}{t}(x-\frac{3t^2}{4})$

$\Rightarrow \frac{2x}{t}-y=\frac{3t}{2}+\frac{t^3}{4}..........\left(2\right)$

Since ${eq}^n\left(1\right)\&\left(2\right)$ represent the same straight line, by comparing the corresponding coefficients, we have

$t=\frac{2}{t}$

$\Rightarrow t=\sqrt{2}$

By substituting the value of 't' in any of the above equaitons, we get

$\sqrt{2}\times x-y=2\sqrt{2}$

The equation of the required straight line is $\sqrt{2}\times x-y=2\sqrt{2}$.