Find the angle of intersection of curve `(x^2)/(a^2)+(y^2)/(b^2)=1` and `x^2+y^2=a b`

Options

A. `"tan"^(-1)((a-b)/(sqrt(ab)))`
B. `"tan"^(-1)((a+b)/(sqrt(ab)))`
C. `"tan"^(-1)((a-b)/(2sqrt(ab)))`
D. none of these

Answer

A

Explanation

We have,
`C_(1):(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 " "...(i) and C_(2):x^(2)+y^(2)=ab " "...(ii)`
Let `P (x_(1),y_(1))` be the point of intersection of these curves. Then, at point P , we have
` ((dy)/(dx))_(C_(1))=-(b^(2)x_(1))/(a^(2)y_(1)) and, ((dy)/(dx))_(C_(2))=-(x_(1))/(y_(1))`
Let ` theta ` be the angle of intersection of (i) and (ii) at point P. Then,
`tan theta =|(((dy)/(dx))_(C_(1))-((dy)/(dx))_(C_(2)))/(1+((dy)/(dx))_(C_(1))((dy)/(dx))_(C_(2)))|`
` rArr tan theta =|((-b^(2)x_(1))/(a^(2)y_(1))+(x_(1))/(y_(1)))/(1+(b^(2)x_(1)^(2))/(a^(2)y_(1)^(2)))|`
` rArr tan theta =((a^(2)-b^(2))x_(1)y_(1))/(a^(2)y_(1)^(2)+b^(2)x_(1)^(2)) `
` rArr tan theta =(a^(2)-b^(2))/(a^(2)b^(2))xx sqrt((a^(3)b^(3))/((a+b)^(2))) [(" Solving (i) and (ii), we have ") , (x_(1)^(2)=(a^(2)b)/(a+b)"," y_(1)^(2)=(ab^(2))/(a+b))]`
` rArr tan theta =((a-b)/(sqrt(ab)))`