Question

The number of different points on the curve $y^2=x(x+1{)}^2$, where the tangent to the curve drawn at (1, 2) meets the curve again, is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Differentiating with respect to y we get
$y^{\prime }=\frac{3x^3+4x+1}{2y}$
Therefore for the slope of the tangent passing through (1,2)
$y^{\prime }=$ $m=$ $\frac{3+4+1}{2\left(2\right)}$ $=2$
Hence equation of tangent line is $y-\left(2\right)=2(x-1)$
$y=2x$
Therefore for the point of intersection
$(2x{)}^2=x(x+1{)}^2$
$x(x-1{)}^2=0$
$x=0$ and $x=1$
Substituting in the equation of the tangent,
$x=0\Rightarrow y=0$
$x=1\Rightarrow y=2$,
But $(1,2)$, is the point at which the tangent is drawn. Hence the tangent meets the curve at one more point.