Question

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

Answer 1

$2y{y}^{\prime }-6x^2-4y^{\prime }=0\Rightarrow y^{\prime }=\frac{3x_1^2}{(y_1-2)}$
Also slope $=\frac{y_1-2}{(x_1-1)}$
Equating
both terms $\frac{3x_1^2}{(y_1-2)}=\frac{(y_1-2)}{(x_1-1)}$
$\Rightarrow 3x{{}_1}^2(x_1-1)={(y_1-2)}^2$
$\Rightarrow 3x{{}_1}^3-3x{{}_1}^2=y{{}_1}^2-4y_1+4$
$\Rightarrow 3x{{}_1}^3-3x{{}_1}^2=(2x{{}_1}^3-8{)}_1+4$
$\Rightarrow x{{}_1}^3-3x{{}_1}^2+4=0$
$\Rightarrow (x_1+1)(x{{}_1}^2-4x_1+4)=0$
$\Rightarrow (x_1+1){(x_1-2)}^2=0$
get the equation of tangent at $x_1=-1,x_1=2$