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प्रश्न
Choose the correct option from the given alternatives :
The normal to the curve x2 + 2xy – 3y2 = 0 at (1, 1)
विकल्प
meets the curve again in second quadrant
does not meet the curve again
meets the curve again in third quadrant
meets the curve again in fourth quadrant
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उत्तर
meets the curve again in fourth quadrant
Explanation:
x2 + 2xy - 3y2 = 0
∴ `2x + 2(xdy/dx + y xx 1) - 3 xx 2ydy/dx` = 0
∴ `(2x - 6y)dy/dx = -2x - 2y`
∴ `dy/dx = (-(x + y))/(x - 3y)`
∴ `(dy/dx)_("at" (1, 1)) = (-1(1 + 1))/(1 - 3)` = 1
= slope of the tangent at (1, 1)
= slope of the normal at (1, 1) is – 1
= equation of the normal is
y – 1
= – 1(x – 1)
= – x + 1
∴ x + y = 2
∴ y = 2 – x
Substituting y = 2 – x in x2 + 2xy – 3y2 = 0, we get
x2 + 2x(2 – x) – 3(2 – x)2 = 0
∴ x2 + 4x – 2x2 – 3(4 – 4x + x2) = 0
∴ x2 – 4x + 3 = 0
∴ (x – 1)(x – 3) = 0
∴ x = 1, x = 3
When x = 1, y = 2 – 1 = 1
When x = 3, y = 2 – 3 = –1
∴ The normal at (1, 1) meets the curve at (3, –1) which is in the fourth quadrant.
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