The portion of the tangent at any point on the curve $x = a t^{3}$ , $y = a t^{4}$ between the axes is divided by the abscissa of the point of contact externally in ratio

$\frac{1}{4}$

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$\frac{3}{4}$

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$\frac{3}{2}$

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$\frac{2}{5}$

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Solution

The correct option is B
$\frac{3}{4}$

$\frac{d y}{d x} = \frac{4 t}{3}$
Tangent is $y - a t^{4} = \frac{4 t}{3} (x - a t^{3})$
x - intercept $= \frac{a t^{3}}{4}$ and y - intercept $= - \frac{a t^{4}}{3}$
The point of intersection of tangent with the axes are $(\frac{a t^{3}}{4}, 0) a n d (0, - \frac{a t^{4}}{3})$
$= \frac{m}{n} = - \frac{3}{4}$
Hence the correct option is (b)

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Similar questions

The portion of the tangent at any point on the curve $x = a t^{3}$ , $y = a t^{4}$ between the axes is divided by the abscissa of the point of contact externally in ratio

Q. The tangent at any point of the curve

x = {a t}^{3}, y = {a t}^{4}

divides the abscissa of the point of contact in the ratio

Q. For the curve

x^{2} y^{3} = C

(where

C

is constant ), the portion of the tangent between the axes is divided by the point of tangents in the ratio of

Q. In the curve

x^{a} y^{b} = k^{a + b}, (a b > 0)

then the portion of the tangent intercepted between coordinate axes is divided at its point of contact into segments which are in constant ratio.

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The portion of the tangent at any point on the curve x=at3, y=at4 between the axes is divided by the abscissa of the point of contact externally in ratio

The correct option is B 34 dydx=4t3 Tangent is y−at4=4t3(x−at3) x - intercept =at34 and y - intercept =−at43 The point of intersection of tangent with the axes are (at34, 0) and (0, −at43) =mn=−34 Hence the correct option is (b)

The portion of the tangent at any point on the curve $x = a t^{3}$ , $y = a t^{4}$ between the axes is divided by the abscissa of the point of contact externally in ratio