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.

Q. The length of the portion of the tangent to the curve

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 4

at any point on it, intercepted between the coordinate axes is constant .

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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