Question

Find the equations of the common tangents to the parabolas $y=x^2-5x+6andy=x^2+x+1.$

Answer 1

$x^2-5x+6=y=f\left(x\right)$

$x^2+x+1=y=g\left(x\right)$

$y=mx+c=tangent$

$f^{\prime }\left(x_1\right)=2x_{1}5=m$

$f^{\prime }\left(x_2\right)=2x_2+1=m$

$2x_1-5.2x_2+1$

$x_1-x_2=3\Rightarrow x_1=x_2+3$

$y_1=x_1^2-5x_1+6$

$y_2=x_2^2+x+1$

$=(x_1-3)+x_1-3+7$

$=x_1^2-5x_1+7$

$y_1=m{x}_1+c$

$\Rightarrow y_1=(2x_1-5)x_1+c$

$\Rightarrow x_1^2-5x_1+6=2x_1^2-5x+c$

$C=6-x_1^2$

also, $y_2=m{x}_2+c$

$x_1^2-5x_1+7=(2x_1-5)(x_1-3)+c$

$x_1^2-5x_1+7=2x_1^2-11x_1+5+6-x_1^2$

$6x_1=14$

$x_1=\frac{7}{3},x_2=\frac{-2}{3}$

$\therefore commontangent=y=\frac{-1}{3}x+\frac{5}{9}$