+0  
 
+1
51
2
avatar

Find the equation of the line that is tangent to both of the parabolas y^2 = 4x and x^2 = -20y. Enter your answer in the form y = mx + b.

 Jun 24, 2022
 #2
avatar+13793 
+1

Find the equation of the line that is tangent to both of the parabolas y^2 = 4x and x^2 = -20y.

 

Hello Guest!

 

\(f(x)=y=\pm2\sqrt{x}\\ f'(x)=\pm\ x^{-\frac{1}{2}}\\ g(x)=y=-\frac{x^2}{20}\\ g'(x)=-\frac{x}{10}\\ f'(x)=g'(x)\\ -\frac{2x}{10}=\pm \ x^{-\frac{1}{2}}\\ \color{blue}x = \pm 4,641588833408518\\ © Arndt\ Br\ddot unner\)

\(y_f=2\cdot \sqrt{4,641588833408518}=4.30886937996\\ P_f(4,641588833408518,4.30886937996)\\ y_g=-\frac{x^2}{20}=-1.07721734492\\ P_g(-4,641588833408518,-1.07721734492)\)

\(m=\dfrac{y_f-y_g}{x_f-x_g}=\dfrac{4.30886937996-(-1.07721734492)}{4,641588833408518-(-4,641588833408518)}\\ \color{blue}m=0.580198604205\)

 

Point-direction equation of the straight line.

\(y=m(x-x_1)+y_1\\ y=0.580198604205(x-4,641588833408518)+4.30886937996\\ The\ equation\ of\ the\ line\ that\ is\ tangent\ to\ both\ of\ the\ parabolas\\ y^2 = 4x\\ and\\ x^2 = -20y\\ is\\\color{blue}y=0.5802x+1.692\)

laugh  !

 Jun 26, 2022
edited by asinus  Jun 26, 2022
edited by asinus  Jun 26, 2022
edited by asinus  Jun 26, 2022
edited by asinus  Jun 26, 2022

15 Online Users