Question

Find the equations of the common tangents to the parabolas $y=x^2$ and $y=-x^2+3x-2.$

Answer 1

Given Parabola: $y=x^2.......\left(i\right)(a_1=\frac{1}{4})$

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

$=>{(x-\frac{3}{2})}^2=-y-2+\frac{9}{4}$

$=>{(x-\frac{3}{2})}^2=-(y-\frac{1}{4}).........\left(ii\right)(a_2=\frac{1}{4})$

Let the equation of common tangent be $y=mx+c..............\left(iii\right)$

as $\left(iii\right)$ is tangent to $\left(i\right)$, $=>c=-a_1{m}^2$

$=>c=-\frac{1}{4}{m}^2...............\left(iv\right)$

as $\left(iii\right)$ also tangent to $\left(ii\right)$, $=>c+mh=k+a{m}^2$

$=>c+m\left(\frac{3}{2}\right)=\frac{1}{4}+\frac{1}{4}{m}^2...........\left(v\right)$

From $\left(iv\right)$ and $\left(v\right)$,

$-\frac{1}{4}{m}^2+m\left(\frac{3}{2}\right)=\frac{1}{4}+\frac{1}{4}{m}^2$

$=>\frac{1}{2}{m}^2-\frac{3m}{2}+\frac{1}{4}=0$

$=>2m^2-6m+1=0$

$=>m=\frac{3+\sqrt{7}}{2},\frac{3-\sqrt{7}}{2}$

And $c=-\frac{1}{4}({\frac{3+\sqrt{7}}{2})}^2$ at $m=\frac{3+\sqrt{7}}{2}$,

$=>c=-\left(\frac{16+6\sqrt{7}}{16}\right)$

And $=>c=-\frac{1}{4}({\frac{3-\sqrt{7}}{2})}^2$ at $m=\frac{3-\sqrt{7}}{2}$

$=>c=\frac{-(16-6\sqrt{7})}{16}$

So, Equation of tangents,

$16y=8(3+\sqrt{7})x-16-6\sqrt{7}$.

And $16y=8(3-\sqrt{7})x-16+6\sqrt{7}$.