Question
The number of tangents to the curve $${y^2} - 2{x^3} - 4y + 8 = 0$$ that pass through $$\left( {1,\,2} \right)$$ is :
A.
$$3$$
B.
$$1$$
C.
$$2$$
D.
$$6$$
Answer :
$$2$$
Solution :
Differentiating w.r.t. $$x,$$
$$2y\frac{{dy}}{{dx}} - 6{x^2} - 4\frac{{dy}}{{dx}} = 0\,\,\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{y - 2}}$$
$$\therefore $$ the equation of the tangent at $$\left( {\alpha ,\,\beta } \right)$$ is $$y - \beta = \frac{{3{\alpha ^2}}}{{\beta - 2}}\left( {x - \alpha } \right)$$
It passes through (1, 2) if $$2 - \beta = \frac{{3{\alpha ^2}}}{{\beta - 2}}\left( {1 - \alpha } \right)$$
or $${\left( {\beta - 2} \right)^2} = 3{\alpha ^2}\left( {\alpha - 1} \right)$$
Also, $$\left( {\alpha ,\,\beta } \right)$$ satisfies the equation of the curve.
$$\eqalign{
& \therefore {\beta ^2} - 2{\alpha ^3} - 4\beta + 8 = 0\,\,\,{\text{or}}\,\,\,{\left( {\beta - 2} \right)^2} = 2{\alpha ^3} - 4 \cr
& \therefore {\left( {\beta - 2} \right)^2} = 3{\alpha ^2}\left( {\alpha - 1} \right) = 2{\alpha ^3} - 4 \cr
& \therefore {\alpha ^3} - 3{\alpha ^2} + 4 = 0 \cr
& {\text{or, }}\left( {\alpha - 2} \right)\left( {{\alpha ^2} - \alpha - 2} \right) = 0 \cr
& {\text{or, }}{\left( {\alpha - 2} \right)^2}\left( {\alpha + 1} \right) = 0 \cr
& {\text{When }}\alpha = 2,\,{\left( {\beta - 2} \right)^2} = 12 \cr
& {\text{or, }}\beta = 2 \pm 2\sqrt 3 \cr
& {\text{When }}\alpha = - 1,\,{\left( {\beta - 2} \right)^2} = - 6 \cr
& {\text{or, }}\beta = {\text{nonreal number}} \cr
& \therefore \,\left( {\alpha ,\,\beta } \right)\,{\text{has two values}}{\text{.}} \cr} $$