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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
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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