The number of those tangents to the curve y^2 | vedclass

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(B) Let $P(\alpha, \beta)$ be any point on the curve $y^2 - 2x^3 - 4y + 8 = 0$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} - 6x^2 -

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$2$

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(B) Let $P(\alpha, \beta)$ be any point on the curve $y^2 - 2x^3 - 4y + 8 = 0$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} - 6x^2 - 4 \frac{dy}{dx} = 0$. $\Rightarrow (2y - 4) \frac{dy}{dx} = 6x^2$. $\Rightarrow \frac{dy}{dx} = \frac{6x^2}{2y - 4} = \frac{3x^2}{y - 2}$. Thus,the slope $m$ at $P(\alpha, \beta)$ is $m = \frac{3\alpha^2}{\beta - 2}$. The equation of the tangent at $P$ is $y - \beta = \frac{3\alpha^2}{\beta - 2} (x - \alpha)$. Since the tangent passes through $(1, 2)$,we have $2 - \beta = \frac{3\alpha^2}{\beta - 2} (1 - \alpha)$. $\Rightarrow -(\beta - 2)^2 = 3\alpha^2 (1 - \alpha)$. $\Rightarrow 3\alpha^3 - 3\alpha^2 - \beta^2 + 4\beta - 4 = 0 \dots (i)$. Also,$P(\alpha, \beta)$ lies on the curve,so $\beta^2 - 4\beta + 8 = 2\alpha^3 \dots (ii)$. Substituting $(ii)$ into $(i)$,we get $3\alpha^3 - 3\alpha^2 - (2\alpha^3 + 4\beta - 8) + 4\beta - 4 = 0$. $\Rightarrow \alpha^3 - 3\alpha^2 + 4 = 0$. Factoring the cubic equation,we get $(\alpha + 1)(\alpha - 2)^2 = 0$. If $\alpha = -1$,then $\beta^2 - 4\beta + 8 = 2(-1)^3 = -2$,so $\beta^2 - 4\beta + 10 = 0$. The discriminant $D = 16 - 40 = -24 If $\alpha = 2$,then $\beta^2 - 4\beta + 8 = 2(2)^3 = 16$,so $\beta^2 - 4\beta - 8 = 0$. Solving for $\beta$,we get $\beta = \frac{4 \pm \sqrt{16 + 32}}{2} = 2 \pm 2\sqrt{3}$. Thus,there are two distinct points of tangency $(2, 2 + 2\sqrt{3})$ and $(2, 2 - 2\sqrt{3})$,which yield two distinct tangents. Therefore,the number of tangents is $2$.

The number of those tangents to the curve $y^2 - 2x^3 - 4y + 8 = 0$ which pass through the point $(1, 2)$ is

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