Question

The slope of the straight line which is both tangent and normal to the curve $4x^3=27y^2$ is

• A. $\underset{–}{+}1$
• B. $\underset{–}{+}\frac{1}{2}$
• C. $\underset{–}{+}\frac{1}{\sqrt{2}}$
• D. $\underset{–}{+}\sqrt{2}$

Answer 1

The correct option is D $\underset{–}{+}\sqrt{2}$

$Given:$
$4x^3=27y^2$
$differentiatingw.r.t.x$
$12x^2=54y{y}^{\prime }$
$t$
$The\text{line is tangent at}\text{P}(3t^2,2t^3)andnormalat(3t_1^2,2t_1^3)$
$\frac{dy}{dx}=\frac{2\left(9t^4\right)}{9\left(2t^3\right)}=1$
$equationto\tan gentP\left(t\right)is$
$y-2t^3=t(x-3t^2)$
$\Rightarrow tx-y=t^3........\left(i\right)$
$EquationtonormalatQ\left(t_1\right)is$
$y-2t_1^3=-\frac{1}{t_1}(x-3t_1^2)$
$\Rightarrow x+t_{1}y=2t_1^4+3t_1^2...........\left(ii\right)$
$\left(i\right)and\left(ii\right)areidentical$
$\frac{t}{1}=\frac{-1}{t_1}=\frac{t^3}{2t_1^4+3t_1^2}$
$\Rightarrow t^2=\frac{2}{t^4}+\frac{3}{t^2}or{t}_1=-\frac{1}{t}$
$\Rightarrow t^6=2+3t^2$
$\Rightarrow t^6-3t^2-2=0$
$\Rightarrow t^2=2$
$t=\pm \sqrt{2}$