1.

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 + dThe 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

Answer»

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.


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