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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Find the joint equation of the pair of a line through the origin and perpendicular to the lines given by x2 + 4xy - 5y2 = 0

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प्रश्न

Find the joint equation of the pair of a line through the origin and perpendicular to the lines given by

x2 + 4xy - 5y2 = 0

बेरीज
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उत्तर

Comparing the equation x2 + 4xy - 5y2 = 0 with ax2 + 2hxy + by2 = 0, we get,

a = 1, 2h = 4, b = -5

Let m1 and m2 be the slopes of the lines represented by x2 + 4xy - 5y2 = 0

∴ m1 + m2 = `(-"2h")/"b" = 4/5`  and  m1m2 = `"a"/"b" = (-1)/5`    ...(1)

Now, required lines are perpendicular to these lines

∴ their slopes are `(-1)/"m"_1` and `1/"m"_2`

Since these lines are passing through the origin, their separate equations are

y = `(-1)/"m"_1 "x"` and y = `(-1)/"m"_2 "x"`

i.e. m1y = - x and m2y = - x

i.e. x + m1y  = 0 and x + m2y = 0

∴ their combined equation is

(x + m1y)(x + m2y) = 0

∴ x2 + (m1 + m2)xy + m1m2y2 = 0

∴ `"x"^2 + 4/5 "xy" - 1/5"y"^2 = 0`   ....[By(1)]

∴ 5x2 + 4xy - y2 = 0

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पाठ 4: Pair of Straight Lines - Miscellaneous Exercise 4 [पृष्ठ १३१]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 4 Pair of Straight Lines
Miscellaneous Exercise 4 | Q 4.1 | पृष्ठ १३१

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