Question

The number of tangents to the curve $y^2-2x^3-4y+8=0$ that pass through $(1,0)$ is

• A. $3$
• B. $1$
• C. $2$
• D. $6$

Answer 1

Answer: (C)

$2$

Point $(0,1)$ doesn't lie on curve, so let $(x_1,y_1)$ is point of contact
Therefore slope of $y^2-2x^3-4y+8=0$ is
$\frac{dy}{dx}=\frac{6x^2}{2y-4}=\frac{6x_1^2}{2y_1-2}$
And slope of line trough $(1,0)$ and $(x_1,y_1)$ is $\frac{0-y_1}{1-x_1}$
Equating slopes we get $6x_1^2-6x_1^3=-2y_1^2-4y_1$
And on solving it with equation of curve we get a quadratic equation. That give two values of slope of tangent.
Hence 2 tangents is possible.