Question Number 100817 by ajfour last updated on 28/Jun/20

Commented byajfour last updated on 28/Jun/20

Find side s of square given a,b,r.

Answered by mr W last updated on 28/Jun/20

right vertex (k,h)  eqn. tangent line to ellipse:  x−y−k+h=0  eqn. tangent line to circle:  x+y−k−h=0  eqn. of circle:  x^2 +(y−b−r)^2 =r^2     (−k+h)^2 =a^2 +b^2  ⇒k−h=(√(a^2 +b^2 ))  (b+r−k−h)^2 =2r^2 ⇒k+h=b+((√2)+1)r  ⇒k=(((√(a^2 +b^2 ))+b+((√2)+1)r)/2)  ⇒h=((b+((√2)+1)r−(√(a^2 +b^2 )))/2)  ⇒s=(√2)k=(((√(a^2 +b^2 ))+b+((√2)+1)r)/(√2))

Commented bymr W last updated on 28/Jun/20

Commented byajfour last updated on 28/Jun/20

(−k+h)^2 =a^2 +b^2  ⇒k−h=(√(a^2 +b^2 ))  some explanation for this line  Sir ....

Commented bymr W last updated on 28/Jun/20

if y=mx+c tangents the ellipse, then  a^2 m^2 +b^2 =c^2 .  i think once you have proved this.

Commented byajfour last updated on 29/Jun/20

Understood Sir, nice and cool!

Answered by ajfour last updated on 29/Jun/20

{(b+r+r(√2))−s(√2) }^2 =a^2 +b^2   ⇒  s=(((√(a^2 +b^2 ))+b+r(1+(√2)))/(√2)) .

Commented byajfour last updated on 29/Jun/20

your idea implemented straightway,  Sir.

Commented bymr W last updated on 29/Jun/20

👍👍👍