If the curves x2/a + y2/b = 1 and x2/c + y2/d = 1 intersect at right angles, then which one is the correct relation?a)b – a = c – db)a + b = c + dc)a – b = c – dd)a – b = c + dCorrect answer is option 'C'. Can you explain this answer?

Question

If the curves x2/a + y2/b = 1 and x2/c + y2/d = 1 intersect at right angles, then which one is the correct relation?a)b &ndash; a = c &ndash; db)a + b = c + dc)a &ndash; b = c &ndash; dd)a &ndash; b = c + dCorrect answer is option 'C'. Can you explain this answer?

Answer

There are two possible correct relations that satisfy the conditions given:

a) (1/a) + (1/c) = 0

b) (1/a) - (1/c) = 0

To see why, we can use the fact that two curves intersect at right angles if and only if the product of their slopes at the point of intersection is -1.

For the first curve, we have:

x^2/a + y^2/b = 1

Differentiating implicitly with respect to x, we get:

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

Solving for dy/dx, we get:

dy/dx = -b(x/y)

At the point of intersection (x,y), the slope of the first curve is:

m1 = -b(x/y)

For the second curve, we have:

x^2/c + y^2/d = 1

Differentiating implicitly with respect to x, we get:

(2x/c) + (2y/d) * (dy/dx) = 0

Solving for dy/dx, we get:

dy/dx = -d(x/y)

At the point of intersection (x,y), the slope of the second curve is:

m2 = -d(x/y)

The product of the slopes is:

m1 * m2 = (-b/c) * (x/y) * (-d(x/y)) = (bd/c)

For the curves to intersect at right angles, we must have:

m1 * m2 = -1

Therefore, we need:

(bd/c) = -1

Solving for c, we get:

c = -bd

Substituting this into the equation for the second curve, we get:

x^2/c + y^2/d = 1

x^2/(-bd) + y^2/d = 1

x^2/d - y^2/b = 0

This is the equation of a hyperbola with transverse axis along the x-axis and conjugate axis along the y-axis. Therefore, the correct relation is:

b) (1/a) - (1/c) = 0

Answer

We have, x²/a + y²/b = 1 ……….(1)
and
x²/c + y²/d = 1 ……….(2)
Let, us assume curves (1) and (2) intersect at (x₁, y₁). Then
x₁²/a + y₁²/b = 1 ……….(3)
and
x₁²/c + y₁²/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 bdx₁² = -acy₁² ……….(5)
Now, (3) – (4) gives,
bdx₁²(c – a) = acy₁²(d – b) ……….(6)
Dividing (6) by (5) we get,
c – a = d – b
Or a – b = c – d.

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