If the curves x^2/a + y^2/b = 1 and x^2/c + y^2/d = 1 intersect at right angles, then which one is the correct relation?

(a) b – a = c – d

(b) a + b = c + d

(c) a – b = c – d

(d) a – b = c + d

The question was posed to me by my school principal while I was bunking the class.

My doubt stems from Calculus Application in division Application of Calculus of Mathematics – Class 12

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Correct answer is (c) a – b = c – d

For explanation: We have, x^2/a + y^2/b = 1 ……….(1) and x^2/c + y^2/d = 1 ……….(2)

Let, us assume curves (1) and (2) intersect at (x1, y1). Then

x1^2/a + y1^2/b = 1 ……….(3) and x1^2/c + y1^2/d = 1 ……….(4)

Differentiating both side of (1) and (2) with respect to x we get,

2x/a + 2y/b(dy/dx) = 0

Or dy/dx = -xb/ya

Let, m1 and m2 be the slopes of the tangents to the curves (1) and (2) respectively at the point (x1, y1); then,

m1 = [dy/dx](x1, y1) = -(bx1/ay1) and m2 = [dy/dx](x1, y1) = -(dx1/cy1)

By question as the curves (1) and (2) intersects at right angle, so, m1m2 = -1

Or -(bx1/ay1)*-(dx1/cy1) = -1

Or bdx1^2 = -acy1^2 ……….(5)

Now, (3) – (4) gives,

bdx1^2(c – a) = acy1^2(d – b) ……….(6)

Dividing (6) by (5) we get,

c – a = d – b

Or a – b = c – d.