Question

If the tangent at $(1,1)$ on $y^2=x(2-x^2)$ meets the curve again at $P$, then $P$ is

• A. $(4,4)$
• B. $(-1,2)$
• C. $(9/4,3/8)$
• D. $Noneofthese$

Answer 1

Answer: (C)

$(9/4,3/8)$

$2y.\frac{dy}{dx}=(2-x{)}^2-2x(2-x)$
At $(1,1)$ we get
$2.\frac{dy}{dx}=1-2$
Or
$\frac{dy}{dx}=\frac{-1}{2}$
Now Equation of the tangent is
$(y-1)=\frac{-1}{2}(x-1)$
Or
$2y-2=-x+1$
Or
$x+2y=3$
Or
$x=3-2y$.
Substituting in the equation we get
$y^2=(3-2y)(2-3+2y{)}^2$
Or
$y^2=(3-2y)(2y-1{)}^2$
Or
$y^2=(3-2y)(4y^2-4y+1)$
Or
$y^2=12y^2-12y+3-8y^3+8y^2-2y$
$y^2=-8y^3+20y^2-14y+3$
Or
$8y^3-19y^2+14y-3=0$
Solving the above cubic, we get
$y=\frac{3}{8},\frac{3}{8}$ and $y=1$
Considering $y=\frac{3}{8}$ we get
$x=3-2y$
$=3-\frac{3}{4}$
$=\frac{9}{4}$
Hence the point is
$(\frac{9}{4},\frac{3}{8})$