=1 On differentiating both sides w.r.t. x, we get ‌
2x
a
+‌
2y
b
⋅‌
dy
dx
‌‌=0 ‌
dy
dx
‌‌=‌
−bx
ay
. . . (i) Also, ‌
x2
c
+‌
y2
d
=1 On differentiating both sides w.r.t. x, we get ‌
2x
c
+‌
2y
d
⋅‌
dy
dx
‌‌=0 ‌
dy
dx
‌‌=‌
−dx
cy
∵ Both the curves intersect each other at 90∘. ∴ Tangents at point of intersection must be perpendicular to each other. ∴ Product of slope of tangents =−1 ‌
−bx
ay
×‌
−dx
cy
=−1‌‌ [from Eqs. (i) and (ii)] ⇒‌‌bdx2=−acy2 ... (iii) Also, on subtracting the equation of given curves, we get (‌
x2
a
+‌
y2
b
−1)−(‌
x2
c
+‌
y2
d
−1)=0 ⇒‌‌x2(‌
1
a
−‌
1
c
)+y2(‌
1
b
−‌
1
d
)=0 or x2(‌
1
a
−‌
1
c
)=−y2(‌
1
b
−‌
1
d
) ... (iv) Dividing Eq. (iii) by Eq. (iv), ‌‌
bd
(‌
1
a
−‌
1
c
)
‌‌=‌
ac
(‌
1
b
−‌
1
d
)
⇒‌‌
bd×ac
c−a
‌‌=‌
ac×bd
d−b
⇒‌c−a‌‌=d−b ‌ or ‌‌c−d‌‌=a−b ‌ or ‌‌a−b‌‌=c−d