Question

If normal at point $P(6,2)$ on the curve $xy-x-2y-2=0$ intersects the curve again at $Q(a,b)$, then the value of $12a+b$ is

• A. $6$
• B. $-8$
• C. $3$
• D. $12$

Answer 1

The correct option is A $6$

$xy-x-2y-2=0\dots \left(1\right)$
Differentiate with respect to $x$
$y+x\frac{dy}{dx}-1-\frac{2dy}{dx}=0$
satisfies the point $(6,2)$
$\frac{dy}{dx}=-\frac{1}{4}$
$\Rightarrow$ slope of normal $=4$
Equation of normal is
$y-2=4(x-6)$
$\Rightarrow y-4x+22=0\dots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get
$x(4x-22)-x-2(4x-22)-2=0$
$\Rightarrow 4x^2-31x+42=0$
$\Rightarrow x=6,\frac{7}{4}$
$x=6$ is point $P$, so $x=\frac{7}{4}$ is point $Q$
Thus point where normal interests the curve again is $Q(\frac{7}{4},-15)$
The value of $12a+b=12\times \frac{7}{4}-15$
$=6$