Find the equation of the common tangent in the first quadrant of the circle `x^2+y^2=16` and the ellipse `x^2/25+y^2/4=1`.Also find the length of the intercept of the tangent between the coordinates axes.

Answer

`y=-(2x)/(sqrt(3))+4sqrt((7)/(3)),(14sqrt(3))/(3)`

Explanation

Let the common tangent to ` x^(2) + y^(2) = 16 and ( x^(2))/(25)+(y^(2))/(4)=1` be
`y= mx +4 sqrt( 1+m^(2))`
`and y= mx + sqrt(25 m^(2)+4)`
since ,Eqs ((i) and (ii) are some tangent .
`therefore 4sqrt(1+m^(2))=sqrt(25m^(2)+4)`
`implies 16(1+m^(2))=25m^(2)+4`
`implies 9m^(2)=12`
`implies m=+- 2//sqrt(3)`
since , tangent is in Ist quadrant .
`therefore mlt0`
`implies m=-2//sqrt(3)`
so , the equation of the common tangent is
`y=-(2x)/(sqrt(3))+4sqrt((7)/(3))`
whcih meets coordinatae axes at `A (2 sqrt(7) ,0) and (0,4,sqrt((7)/(3)))`
`therefore AB= sqrt((2sqrt(7)-0)^(2)+(0-4sqrt((7)/(3)))^(2))`
`= sqrt(28+(11)/(3))=sqrt((196)/(3))=(14)/(sqrt(3))xx(sqrt(3))/(sqrt(3))=(14sqrt(3))/(3)`