The number of tangent to curve
$$y^2 – 2x^3 – 4y + 8 = 0$$ that passes through $(1,2)$
My work
Assuming tangent touch the curve at$(x_1,y_1)$
$$\frac{dy}{dx}=\frac{3x^2}{(y – 2)} $$
$$\frac{2 – y_1}{1- x_1}=\frac{dy}{dx}=\frac{3x_1^2}{(y_1 – 2)}$$,
I should need one more equation. How I will get that?
Best Answer
Cross-multiply and you will have:
$$(2-y_1)(y_1 - 2) = 3x_1^2(1-x_1) \implies -y^2 + 4y_1 -4 = 3x_1^2 - 3x_1^3$$
Now substitute $y_1^2 - 2x_1^3 - 4y_1 + 8 =0$ and eventually you will get:
$$-2x_1^3 + 4 = 3x_1^2 - 3x_1^3 \implies x_1^3 - 3x_1^2 + 4 = (x_1-2)^2(x_1+1) = 0$$
But it's fairly easy to see that $x=-1$ doesn't give solutions and you will get 2 distinct solutions from $x=2$