Question

The tangent to the curve $3x{y}^2-2x^{2}y=1$ at $(1,1)$ meets the curve again at the point

• A. $(\frac{16}{5},\frac{1}{20})$
• B. $(-\frac{16}{5},-\frac{1}{20})$
• C. $(\frac{1}{20},\frac{16}{5})$
• D. $(-\frac{1}{20},\frac{16}{5})$

Answer 1

The correct option is B $(-\frac{16}{5},-\frac{1}{20})$

Given, $3x{y}^2-2x^{2}y=1$

Differentiating with respect to x, gives us

$3y^2+6xy{y}^{\prime }-4xy-2x^2{y}^{\prime }=0$

At (1,1)

$3+6y^{\prime }-4-2y^{\prime }=0$

$4y^{\prime }=1$

$y^{\prime }=\frac{1}{4}$

Hence the equation of the tangent is

$\frac{y-1}{x-1}=\frac{1}{4}$

$4y-4=x-1$

$x-4y=-3$

$x=4y-3$

Substituting in the equation of the curve, give us

$3(4y-3)y^2-2(4y-3{)}^{2}y=1$

$(4y-3)y[3y-2(4y-3\left)\right]=1$

$(4y-3)y(6-5y)=1$

$y(4y-3)(5y-6)=-1$

$y(20y^2-24y-15y+18)=-1$

$y(20y^2-39y+18)=-1$

$20y^3-39y^2+18y+1=0$

$y=1,\frac{-1}{20}$

$\Rightarrow x=1,\frac{-16}{5}$

Hence the tangent again cuts the curve at $(\frac{-16}{5},\frac{-1}{20})$