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

$\frac{dy}{dx}=2x-1$
Slope of the normal,
$m=\frac{1}{1-2x}$

At $x_1=0\Rightarrow y=1$
$\Rightarrow m_1=1$
Equation of the normal:
$y-1=1(x-0)$
$\Rightarrow x-y+1=0...\left(1\right)$

At $x_2=-1\Rightarrow y=3$
$\Rightarrow m_2=\frac{1}{3}$
Equation of the normal:
$y-3=\frac{1}{3}(x+1)$
$\Rightarrow x-3y+10=0...\left(2\right)$

At $x_3=\frac{5}{2}\Rightarrow y=\frac{19}{4}$
$\Rightarrow m_3=-\frac{1}{4}$
Equation of the normal:
$y-\frac{19}{4}=-\frac{1}{4}(x-\frac{5}{2})$
$\Rightarrow 2x+8y-43=0...\left(3\right)$

$\Delta =\left|\begin{array}{ccc}1& -1& 1\\ 1& -3& 10\\ 2& 8& -43\end{array}\right|$
$=0$
Therefore, normals are concurrent.