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(D) Given the curve $y = x^{3} + 3x^{2} + 5$. Let the point of tangency be $(x_{1}, y_{1})$. The slope of the tangent at $(x_{1}, y_{1})$ is $\frac{dy

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the curve $y = x^{3} + 3x^{2} + 5$. Let the point of tangency be $(x_{1}, y_{1})$. The slope of the tangent at $(x_{1}, y_{1})$ is $\frac{dy}{dx} = 3x^{2} + 6x$. So,the slope at $(x_{1}, y_{1})$ is $m = 3x_{1}^{2} + 6x_{1}$. The equation of the tangent at $(x_{1}, y_{1})$ is $y - y_{1} = (3x_{1}^{2} + 6x_{1})(x - x_{1})$. Since the tangent passes through the origin $(0, 0)$,we have: $0 - y_{1} = (3x_{1}^{2} + 6x_{1})(0 - x_{1})$ $-y_{1} = -3x_{1}^{3} - 6x_{1}^{2} \implies y_{1} = 3x_{1}^{3} + 6x_{1}^{2} \quad (1)$. Since $(x_{1}, y_{1})$ lies on the curve $y = x^{3} + 3x^{2} + 5$,we have: $y_{1} = x_{1}^{3} + 3x_{1}^{2} + 5 \quad (2)$. Equating $(1)$ and $(2)$: $3x_{1}^{3} + 6x_{1}^{2} = x_{1}^{3} + 3x_{1}^{2} + 5$ $2x_{1}^{3} + 3x_{1}^{2} - 5 = 0$. Testing for roots,$x_{1} = 1$ is a root. Dividing by $(x_{1} - 1)$,we get $(x_{1} - 1)(2x_{1}^{2} + 5x_{1} + 5) = 0$. For $x_{1} = 1$,$y_{1} = 3(1)^{3} + 6(1)^{2} = 9$. Thus,the point is $(1, 9)$. Checking which curve does not contain $(1, 9)$: For $D$: $\frac{1}{3} - (9)^{2} = \frac{1}{3} - 81 
eq 2$. Therefore,$(1, 9)$ does not lie on the curve $\frac{x}{3} - y^{2} = 2$.

If the tangent at the point $(x_{1}, y_{1})$ on the curve $y=x^{3}+3x^{2}+5$ passes through the origin,then $(x_{1}, y_{1})$ does $NOT$ lie on which of the following curves?

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