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प्रश्न
If the line y = 4x – 5 touches the curves y2 = ax3 + b at the point (2, 3), find a and b.
उत्तर
y2 = ax3 + b
Differentiating both sides w.r.t. x, we get
`2ydy/dx = a xx 3x^2 + 0`
∴ `dy/dx = (3ax^2)/(2y)`
∴ `(dy/dx)_("at" (2, 3)) = (3a(2)^2)/(2(3))`
= 2a
= slope of the tangent at (2, 3)
Since, the line y = 4x – 5 touches the curve at the point (2, 3), slope of the tangent at (2, 3) is 4.
∴ 2a = 4
∴ a = 2
Since (2, 3) lies on the curve y2 = ax3 + b,
(3)2 = a(2)3 + b
∴ 9 = 8a + b
∴ 9 = 8(2) + b ...[∵ a = 2]
∴ b = – 7
Hence, a = 2 and b = – 7.
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