Question

Find the tangents and normal to the curve $y(x-2)(x-3)-x+7=0,$ at point (7,0) are

• A. $x-20y-7=0,$ $20x+y-140=0$.
• B. $x+20y-7=0,$ $20x-y-140=0$.
• C. $7x-20y-1=0,$ $20x+7y-100=0$.
• D. $7x+20y-1=0,$ $20x-7y-100=0$.

Answer 1

The correct option is A $x-20y-7=0,$ $20x+y-140=0$.

Given curve is, $y(x-2)(x-3)-x+7=0\Rightarrow y(x^2-5x+6)-x+7=0$
Differentiating w.r.t $x$
$\frac{dy}{dx}(x^2-5x+6)+y(2x-5)-1=0$
putting (7,0) to get slope of the tangent
$\frac{dy}{dx}(7^2-5.7+6)+0(2x-5)-1=0\Rightarrow \frac{dy}{dx}=\frac{1}{20}$
$\Rightarrow$ slope of tangent is $m=\frac{1}{20}$ and slope of normal is, $m^{\prime }=-\frac{1}{m}=-20$
Hence required lines are $x-20y-7=0,$ and $20x+y-140=0$.