Question

If the tangent at $(1,1)$ on $y^2=x(2-x{)}^2$ meets the curve again at $P(a,b)$ then $a/b$ is equal to

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (C)

$6$

$y^2=x(2-x{)}^2$
$\Rightarrow 2y{y}^{\prime }=(2-x{)}^2+2x(x-2)$
$\Rightarrow y^{\prime }=\frac{3x^2-8x+4}{2y}\Rightarrow y_{(1,1)}^{\prime }=-\frac{1}{2}$
$\Rightarrow$ Equation of tangent is $y-1=-\frac{1}{2}(x-1)$
$\Rightarrow 2y-2=1-x\Rightarrow x+2y=3$
Solve with $y^2=x(x-2{)}^2$
$\Rightarrow (x-3{)}^2=4x(x-2{)}^2$
$\Rightarrow x^2-6x+9=4x^3-16x^2+16x$
$\Rightarrow 4x^3-17x^2+22x-9=0$
$\Rightarrow (x-1{)}^2(4x-9)=0$
$\Rightarrow x=\frac{9}{4}\Rightarrow y=\frac{3}{8}$
$\Rightarrow \frac{a}{b}=\frac{9\times 8}{4\times 3}=6$