Question

Let $P(x_0,y_0)$ be a point on the curve $C:(x^2-11)(y+1)+4=0,$ where $x_0,y_0\in N.$ If area of the triangle formed by the normal drawn to the curve $C$ at $P$ and the co-ordinate axes is $\frac{a}{b},$ where $a,b\in N,$ then the least value of $a-6b$ is

Answer 1

$(x^2-11)(y+1)+4=0$
$\Rightarrow (x^2-11)(y+1)=-4=-2\times 2$
On comparing, we get
$P(x_0,y_0)\equiv (3,1)$ as $x_0,y_0\in N$
$y^{\prime }=\frac{8x}{(x^2-11{)}^2}$
$\Rightarrow y^{\prime }{|}_{x=3}=6$
$\therefore$ Slope of normal, $m_N=-\frac{1}{6}$

Equation of normal at $P(3,1)$ is
$y-1=-\frac{1}{6}(x-3)$
$\Rightarrow x+6y=9$


Area $=\frac{1}{2}\times 9\times \frac{3}{2}=\frac{27}{4}$