Question

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

Answer 1

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

Equation of tangent at $(2,0)$

$\Rightarrow \frac{y+0}{2}=2x-\frac{-5(x+2)}{2}+6$

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

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

$\Rightarrow y+x+4=0$

$\Rightarrow x+y+4=0$
$\Rightarrow y=-x-4\cdot \cdot \cdot \left(1\right)$
Slope of line $=m_1=-1$

Equation of tangent at$(3,0)$

$\Rightarrow \frac{y+0}{2}=3x-\frac{-5(x+3)}{2}+6$

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

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

$\Rightarrow y=x-9\cdot \cdot \cdot \left(2\right)$
Slope of line $=m_2=1$

from equation (1) &(2)

$m_1\times m_2=-1$
here product of slope of tangent is $-1.$

$\therefore \text{Angle between tangent}={90}^{\circ }$
Hence, the correct answer is Option a.