The tangent at (x₁, y₁) on the curve $y = x^{3} + 3 x^{2} + 5$ passes through the origin then the point (x₁, y₁) does not lie on the curve

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$x^{2} + \frac{y^{2}}{81} = 2$

$\frac{y^{2}}{9} - x^{2} = 8$

$y = 4 x^{2} + 5$

$\frac{x}{3} - y^{2} = 2$

answer is D.

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Detailed Solution

$y^{3} = x^{3} + 3 x^{2} + 5; P (x_{1}, y_{1}) \frac{d y}{d x} = 3 x^{2} + 6 x \Rightarrow m = 3 {x_{1}}^{2} + 6 x_{1} T h e e q u a t i o n o f t h e \tan g e n t a t p i s y - y_{1} = (3 x_{1}^{2} + 6 x_{1}) (x - x_{1}) I t p a s s e s t h r o u g h o r i g i n \begin{array}{l} 0 - y_{1} = (3 x_{1}^{2} + 6 x_{1}) (0 - x_{1}) \\ \Rightarrow x_{1}^{3} + 3 x_{1}^{2} + 5 = 3 x_{1}^{3} + 6 x_{1}^{2} \\ \Rightarrow 2 x_{1}^{3} + 3 x_{1}^{2} - 5 = 0 \\ x_{1} = 1; y_{1} = 9 \end{array} T h e p o i n t (x_{1}, y_{1}) = (1, 9) d o e s n o t l i e s o n \frac{x}{3} - y^{2} = 2$