Find the equations of the tangents drawn to the curve $y^{2} - {2 x}^{3} - 4 x + 8 = 0$ from the point (1,2)

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Solution

We have,

$y^{2} - 2 x^{3} - 4 x + 8 = 0$

On differentiation and we get,

$2 y \frac{d y}{d x} - 6 x^{2} - 4 + 0 = 0$

$\frac{d y}{d x} = \frac{6 x^{2} + 4}{2}$

$\frac{d y}{d x} = 3 x^{2} + 2$

Slope at the point $(1, 2)$

Then,

${(\frac{d y}{d x})}_{(1, 2)} = 3 {(1)}^{2} + 2$

${(\frac{d y}{d x})}_{(1, 2)} = 5$

Then,

Equation of tangent is

$y - y_{1} = \frac{d y}{d x} (x - x_{1})$

$\Rightarrow y - 2 = 5 (x - 1)$

$\Rightarrow y - 2 = 5 x - 5$

$\Rightarrow y - 5 x + 3 = 0$

Hence, this is the answer.

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