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प्रश्न
Solve the following : If the curves ax2 + by2 = 1 and a'x2 + b'y2 = 1, intersect orthogonally, then prove that `(1)/a - (1)/b = (1)/a' - (1)/b'`.
उत्तर
Let P(x1, y1) be the point of intersection of the curves.
∴ ax12 + by12 = 1 and a'x12 + b'y12 = 1
∴ ax12 + by12 = a'x12 + b'y12
∴ (a – a')x12 = (b' – b)y12 ...(1)
Differentiating ax2 + by2 = 1 w.r.t. x, we get
`a xx 2x + b xx 2y dy/dx = 0`
∴ `dy/dx = (-ax)/(by)`
∴ slope of the tangent at (x1, y1) = m1
= `(dy/dx)_("at" (x_1, y_1)) = (-a'x_1)/(b'y_1)`
Since. curves intersect each other orthogonally,
m1m2 = – 1
∴ `((-ax_1)/(by_1))((-a'x_1)/(b'y_1))` = – 1
∴`(aa'x_1^2)/(b b'y_1^2)` = – 1
∴ `(aa')/(b b') = (-y_1^2)/(x_1^2)`
∴ `(aa')/(b b') = ((a - a')/(b - b'))` ...[By (1)]
∴ `(a - a')/(aa') = (a - a')/(b b')`
∴ `(1)/a' - (1)/a = (1)/b' - (1)/b`
∴ `(1)/a - (1)/b = (1)/a' - (1)/b'`.
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