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Tangent of a Curve y =f(x)

Let C be the ...

Question

Let C be the curve $y = x^{3}$ (where x takes all real values). The tangent at a point A meets the curve again at B. If the gradient at B is K times the gradient at A then K is equal to

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-2

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$\frac{1}{4}$

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Solution

The correct option is A
4

$\frac{d y}{d x} = 3 x^{2} = 3 t^{2} a t^{'} A^{'}$

$∴ 3 t^{2} = \frac{T^{3} - t^{3}}{T - t} = T^{2} + T t + t^{2} \Rightarrow T^{2} + T t - 2 t^{2} = 0 \Rightarrow (T - r) (T + 2 t) = 0 \Rightarrow T = t o r T = - 2 t$ (T = t is not possible)
now, $m_{A} = 3 t^{2}; m_{B} = 3 T^{2}$
$\Rightarrow \frac{m_{B}}{m_{A}} = \frac{T^{2}}{t^{2}} = \frac{4 t^{2}}{t^{2}} (u s i n g T = - 2 t)$
Hence the correct option is (a)

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