Question

State true or false.

The curves $y=x^2-3x+1$ and $x\left(y+3\right)=4$ intersect at the right angles at their point of intersection

• A. True
• B. False

Answer 1

The correct option is A True

First curve is given by,

$y=x^2-3x+1$ (1)

Differentiate w.r.t. x, we get,

$\frac{dy}{dx}=2x-3$

Thus, slope of tangent to this curve is,

$m_1=2x-3$

Second curve is given by,

$x(y+3)=4$

$\therefore (y+3)=\frac{4}{x}$ (2)

Differentiate w.r.t. x, we get,

$\frac{dy}{dx}=-\frac{4}{x^2}$

Thus, slope of tangent to this curve is,

$m_2=-\frac{4}{x^2}$

1) Point of intersection of two curves-

Equation of first curve is given by,

$y=x^2-3x+1$

Equation of second curve is given by,

$y=\frac{4}{x}-3$

Thus, we can equate these equations, we get,

$x^2-3x+1=\frac{4}{x}-3$

Multiply both sides by x, we get,

$\therefore x^3-3x^2+x=4-3x$

$\therefore x^3-3x^2+4x-4=0$

Put $x=2$ in above equation,

$LHS={\left(2\right)}^3-3{\left(2\right)}^2+4\left(2\right)-4$

$=8-(3\times 4)+8-4$

$=8-12+8-4$

$=0=RHS$

Thus, $x=2$ is one factor of given equation.

By synthetic division, we can get quadratic equation as,

$x^2-x+2=0$

$\therefore x=\frac{-(-1)\pm \sqrt{{(-1)}^2-4\times 1\times 2}}{2\times 1}$

$\therefore x=\frac{1\pm \sqrt{1-8}}{2}$

$\therefore x=\frac{1\pm \sqrt{-7}}{2}$

These roots will be imaginary and are neglected.

$\therefore x=2$

Put this value in equation (1), we get,

$y={\left(2\right)}^2-3\left(2\right)+1$

$\therefore y=4-6+1$

$\therefore y=-1$

Thus, point of intersection of two curves is, $(2,-1)$

${\left[m_1\right]}_{(2,-1)}=2\left(2\right)-3$

$\therefore {\left[m_1\right]}_{(2,-1)}=1$

${\left[m_2\right]}_{(2,-1)}=\frac{-4}{{\left(2\right)}^2}$

$\therefore {\left[m_2\right]}_{(2,-1)}=\frac{-4}{4}=-1$

Now, $m_1\times m_2=1\times -1$

$\therefore m_1\times m_2=-1$

Thus, curves intersect each other at right angle