Question

What is the angle between the tangents to the curve $y=x^2-5x+6$ at the points (2, 0) and (3, 0)

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A

$\frac{\pi }{2}$

We have to use the equation $T_1=0$
for getting the tangent at the point $(x_1,y_1)$ situated on the curve.

The required tangent is,

$T_1=0$

i.e., $\frac{y+y_1}{2}=x{x}_1-\frac{-5(x+x_1)}{2}+6$

For (2,0)

$T_1=0$

$\frac{y}{2}=2x\frac{-5}{2}(x+2)+6$

$y=4x-5x-10+6$

$y+x+4=0$

$\Rightarrow x+y+4=0$ .....(1)

For (3,0)

$T_1=0$

$\frac{y}{2}=3x-\frac{5}{2}(x+3)+6$

$y=6x-5x-15+6$

x-y-9=0 .............(2)

from equation (1) &(2)

$m_1\times m_2=-1$

$\therefore Anglebetweentangent={90}^{\circ }$