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


If $(x_1,y_1)$ is a point on the curve $$2x_1^3=y_1^2-4y_1+8\ \ \ \ (1)$$

The equation of the tangent at $(x_1,y_1)$ $$\dfrac{y-y_1}{x-x_1}=\dfrac{6x_1^2}{2y_1-4}$$

Now as it has to pass through $(1,2)$

$$\dfrac{2-y_1}{1-x_1}=\dfrac{6x_1^2}{2y_1-4}\ \ \ \ (2)$$ I am stuck here.The tangents given in the answer are $2\sqrt3x-y=2(\sqrt3-1)$ and $2\sqrt3x+y=2(\sqrt3+1)$


At this point it's basically solving a system of two nonlinear equations in two variables. The equations are what you labeled $(1)$ and $(2)$.

From $(2)$ we can remove a factor of $2$ on the right and clear denominators to get: $$-y_1^2 + 4y_1 - 4 = 3x_1^2 - 3x_1^3$$ And we want to solve this simultaneously with: $$2x_1^3 = y_1^2 - 4y_1 + 8$$ In the first equation above, let's rewrite the LHS as $$-y_1^2 + 4y_1 - 4 = -(y_1^2 - 4y_1 + 8) + 4 = -2x_1^3 + 4.$$ Now solve that first equation above for $x_1$, using $-2x_1^3 + 4$ as the new LHS.

Can you take it from here?


You need to solve for $(x_1, y_1)$ using the combination of equations $(1)$ and $(2)$. I'm not sure how you got Equation $(2)$ from the previous equation, since they are not the same.


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