If the curves `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(c^(2))+(y^(2))/(d^(2))=1` intersect orthogonally, then

Options

A. `a^(2)-b^(2)=c^(2)-d^(2)`
B. `a^(2)-c^(2)=b^(2)-d^(2)`
C. `a^(2)b^(2)=c^(2)d^(2)`
D. `(1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))+(1)/(d^(2))`

Answer

A

Explanation

The equations of the two curves are
`C_(1) : (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 " " `...(i)
` C_(2) :(x^(2))/(c^(2))+(y^(2))/(d^(2))=1 " " ` ...(ii)
Suppose these two curves intersect at `P(x_(1), y_(1)).` Then,
` (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 " " `...(iii)
` and, (x^(2))/(c^(2))+(y^(2))/(d^(2))=1 " " ` ...(iv)
Differentiating (i) and (ii) with respect to x, we get
`((dy)/(dx))_(C_(1))=(b^(2)x)/(a^(2)y) and ((dy)/(dx))_(C_(2))=(d^(2)x)/(c^(2)y)`
If the two curves intersect orthogonally at P. Then,
`-(b^(2)x_(1))/(a^(2)y_(1)) xx (d^(2)x_(1))/(c^(2)y_(1))= -1 rArr b^(2)d^(2)x_(1)^(2)=-a^(2)c^(2)y_(1)^(2) " " `...(v)
Subtracting (iv) from (iii), we get
`(x_(1)^(2))/(a^(2))-(x_(1)^(2))/(c^(2))+(y_(1)^(2))/(b^(2))-(y_(1)^(2))/(d^(2))=0`
`rArr x_(1)^(2)((a^(2)-c^(2)))/(a^(2)c^(2))=y_(1)^(2)((d^(2)-b^(2)))/(b^(2)d^(2))`
`rArr b^(2)d^(2)x_(1)^(2)(a^(2)-c^(2))=a^(2)c^(2)y_(1)^(2)(d^(2)-b^(2))`
`rArr a^(2)-c^(2)= -(d^(2)-b^(2)) " " `[Using (v)]
` rArr a^(2)-b^(2)=c^(2)-d^(2)`