Question

The tangent at any point of the curve $x={at}^3,y={at}^4$ divides the abscissa of the point of contact in the ratio

• A. $1:4$
• B. $3:2$
• C. $1:3$
• D. $3:1$

Answer 1

The correct option is A $1:4$

Given:

$x=a{t}^2$

$y=a{t}^4$

Then,

$\Rightarrow t^3=\frac{x}{a},t^4=\frac{y}{a}$

$\Rightarrow t={\left(\frac{x}{a}\right)}^{1/3}t={\left(\frac{y}{a}\right)}^{1/4}$

$\Rightarrow {\left(\frac{x}{a}\right)}^{1/3}={\left(\frac{y}{a}\right)}^{1/4}$

$\Rightarrow {\left(\frac{x}{a}\right)}^{\frac{1}{3}\times \frac{4}{12}}={\left(\frac{y}{a}\right)}^{\frac{1}{4}\times 12}$

$\Rightarrow \frac{x^4}{a^4}=\frac{y^3}{a^2}$

$\Rightarrow y^3=\frac{x^4}{a}$ at $P(h,k)$

Also relation $k^3=\frac{h^4}{a}$

Now by differentiation of after equation we get

$3y^2\frac{dy}{dx}=\frac{4x^2}{a}$

$\frac{dy}{dx}=\frac{4x^3}{3a{y}^2}$

Now $M_T=\frac{dy}{dx}=\frac{4x^3}{3a{y}^2}$

Now we will find slope of tangent

$M_T/P(h,x)=\frac{4h^3}{3a{x}^2}$

equation of tangent $P(h,k)$

$y-k=\frac{4h^3}{3a{k}^2}(x-h)$

$\Rightarrow$ Let $y=0$

Then, $-k=\frac{4h^3}{3a{k}^2}(x-h)$

$\Rightarrow \frac{-3a{k}^3}{4h^3}=x-h$

Earlier we found that $k^3=\frac{h^4}{a}$

Then $a{x}^3=h^4$

$\frac{-3h^4}{4h^2}=x-h$

$\Rightarrow \frac{-3h}{4}+h=x$

$\Rightarrow x=\frac{h}{4}$

$\Rightarrow \frac{x}{h}=\frac{1}{4}$

Hence option (a) $1:4$ is the correct choice