Question

The normals to the curve $y=x^2-x+1$, drawn at the points with the abscissa $x_1=0,x_2=-1$ and $x_3=\frac{5}{2}$

• A. are parallel to each other
• B. are pair wise perpendicular
• C. are concurrent
• D. are not concurrent

Answer 1

The correct option is C are concurrent

$\frac{dy}{dx}=2x-1,m_N=\frac{1}{1-2x}$
$m_{x_1}=1$ point $(0,1)$
$y-1=1\left(x\right),x-y+1=\mathrm{0.....}\left(1\right)$
$m_{x_2}=\frac{1}{3}$ point $(-1,3)$
$y-3=\frac{1}{3}(x+1)\Rightarrow 3y-9=x+1$
$x-3y+10=\mathrm{0...}\left(2\right)$
$x_3=\frac{5}{2},(\frac{5}{2},\frac{19}{4})$
$m_{x_3}=-\frac{1}{4}$
$y-\frac{19}{4}=-\frac{1}{4}(x-\frac{5}{2})$
$\Rightarrow x+4y=\frac{43}{2}....\left(3\right)$
Intersection point of (1) and (2) $(\frac{7}{2},\frac{9}{12})$ passes (3)
Hence normal are concurrent