Question

The gradient of the common tangent to the two curves $y=x^2-5x+6$ and $y=x^2+x+1$ is

• A. $\frac{-1}{3}$
• B. $\frac{-2}{3}$
• C. $-1$
• D. $-3$

Answer 1

Answer: (D)

$-3$

$y=x^2-5x+6$
$y=x^2-2x\left(\frac{5}{2}\right)+\frac{25}{4}+6-\frac{25}{4}$
$y={(x-\frac{5}{2})}^2-\frac{1}{4}$
$(y+\frac{1}{4})={(x-\frac{5}{2})}^2$
Equation of tangent to $x^2=4ay$ is $x=my+\frac{a}{m}$
$\therefore x-\frac{5}{2}=m[y+\frac{1}{4}]+\frac{1}{4m}⟶1$
$x-\frac{5}{2}-my-\frac{m}{4}-\frac{1}{4m}=0⟶2$
$x^2+x+1=y$
$x^2+x+\frac{1}{4}+1-\frac{1}{4}=y$
${(x+\frac{1}{2})}^2=y-\frac{3}{4}$
Equation of tangent $\Rightarrow x+\frac{1}{2}=m(y-\frac{3}{4})+\frac{1}{4m}$
$x+\frac{1}{2}-my+\frac{3}{4}m-\frac{1}{4m}=0⟶3$
Equation 2 and 3 represents same line
$\therefore x-\frac{5}{2}-my-\frac{m}{4}-\frac{1}{4m}=x+\frac{1}{2}-my+\frac{3}{4}m-\frac{1}{4m}$
$-\frac{m}{4}-\frac{3m}{4}=\frac{1}{2}+\frac{5}{2}$
$-m=\frac{6}{2}$
$m=-3$