Question

The number of real tangents that can be drawn to the curve $y^2+2xy+x^2+2x+3y+1=0$ from the point $(1,-2)$ is

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

Answer: (C)

$0$

Equation of tangent to the curve $S=0$ at any point is $T=0$

Let the tangent from $(-1,2)$ touch the curve at $(x_1,y_1)$

Equation of tangent at $(x_1,y_1)$ is

$y{y}_1+2\left(\frac{x{y}_1+y{x}_1}{2}\right)+x{x}_1+2\left(\frac{x+x_1}{2}\right)+3\left(\frac{y+y_1}{2}\right)+1=0$

This line passes through $(1,-2)$

$⟹-2y_1+2\left(\frac{y_1-2x_1}{2}\right)+x_1+2\left(\frac{1+x_1}{2}\right)+3\left(\frac{-2+y_1}{2}\right)+1=0$

$⟹\frac{1}{2}{y}_1=1$

$⟹y_1=2$

Point $(x_1,y_1)$ lies on the curve

$\therefore y_1^2+2x_1{y}_1+x_1^2+2x_1+3y_1+1=0$

$\therefore 4+4x_1+x_1^2+2x_1+6+1=0$

$\therefore x_1^2+6x_1+11=0$

$b^2-4ac=36-44<0$

Hence, no solution

So, no tangent can be drawn from the point $(1,-2)$

The answer is option (C).