Question

If the slope of the tangent to the curve $x^2-2xy+4y=0$ at a point on it is $\frac{-3}{2}$, then find the equations of tangent and normal at that point.

Answer 1

Let the point be $(a,b)$

$a^2-2ab+4b=0$......(1)

also $dy$ $=\frac{-3}{2}$

$dx$ $(a,b)$

$x^2-2xy+4b=0$

differentiating w.r.t $x$ in both sides

$2x-2y-2x\underset{dx}{dy}+4\underset{dx}{dy}=0$

$x-y-x\underset{dx}{dy}+2\underset{dx}{dy}=0\Rightarrow \frac{x-y}{x-z}=\frac{dy}{dx}$

dy $=\frac{a-b}{a-2}=\frac{-3}{2}\Rightarrow 2a-2b=-3a+6$

dx (a,b)

$5a-2b=6\Rightarrow b=\frac{5a-6}{2}$

substituting value b in $e{q}^n$......(1)

$a^2-7a\left(\frac{5a-6}{2}\right)+4^2\left(\frac{5a-6}{2}\right)=0$

$a^2-5a^2+6a+10a-12=0$

$-4a^2+16a-12=0$

$4a^2-16a+12=0\Rightarrow a^2-4a+3=0$

$a^2-3a-a+3=0$

$a(a-3)-1(a-3)=0$

$(a-1)(a-3)=0\Rightarrow a=1$ or $a=3$

$a=1,b=\frac{5a-6}{2}=\frac{5-6}{2}=-\frac{1}{2}$

$(1,\frac{-1}{2})$

If $a=3$, $b=\frac{5.3-6}{2}=\frac{15-6}{2}=\frac{9}{2}$

$at(1,\frac{-1}{2}$

equation of tanget $\to$

$(y-b)=dy$ $(x-a)$

$dx$ $(a,b)$

$(y+\frac{9}{2}=\frac{-3}{2}(x-1\left)\right)$\

$2y+1=-3x+3$

$3x+2y=2$

equation of normal $\to$

$(y-b)=-dx$ $(x-a)$

$dy$ $(a-b)$

$y+\frac{1}{2}=\frac{2}{3}(x-1)$

$3y+\frac{3}{2}=2x-2$

$3y-2x+\frac{7}{2}=0$

$at3,\frac{9}{2}$

equation at tangent $\to$

$(y-b)=dy$ $(x-a)$

$dx$ $(a,b)$

$(y-\frac{9}{2}=\frac{-3}{2}(x-9)$

$2y-9=-3x+9$

$3x+2y=18$

equation of normal $\to$

$(y-b)=-dx(x-a)$

$dy(a,b)$

$(y-\frac{9}{2}=\frac{2}{3}(x-3)$

$3y-\frac{27}{2}=2x-6$

$3y-2x=\frac{15}{2}$