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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Tangents are drawn through a point P to the ellipse 4x2 + 5y2 = 20 having inclinations θ1 and θ2 such that tan θ1 + tan θ2 = 2. Find the equation of the locus of P.

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प्रश्न

Tangents are drawn through a point P to the ellipse 4x2 + 5y2 = 20 having inclinations θ1 and θ2 such that tan θ1 + tan θ2 = 2. Find the equation of the locus of P.

बेरीज
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उत्तर

Given equation of the ellipse is 4x2 + 5y2 = 20.

∴ `x^2/5 + y^2/4` = 1

Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get

a2 = 5 and b2 = 4

Since inclinations of tangents are θ1 and θ2 ,

m1 = tan θ1 and m2 = tan θ2.

Equation of tangents to the ellipse

`x^2/"a"^2 + y^2/"b"^2` = 1 having slope m are

y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)`

∴ y = `"m"x ± sqrt(5"m"^2 + 4)`

∴ y – mx = `± sqrt(5"m"^2 + 4)`

Squaring both the sides, we get

y2 – 2mxy + m2x2 = 5m2 + 4

∴ (x2 – 5)m2 – 2xym + (y2 – 4) = 0

The roots m1 and m2 of this quadratic equation are the slopes of the tangents.

∴ m1 + m2 = `(-(-2xy))/(x^2 - 5) = (2xy)/(x^2 - 5)`

Given, tan θ1 + tan θ2 = 2

∴ m1 + m2 = 2

∴ `(2xy)/(x^2 - 5)` = 2

∴ xy = x2 – 5

∴ x2 – xy – 5 = 0, which is the required equation of the locus of P.

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पाठ 7: Conic Sections - Exercise 7.2 [पृष्ठ १६४]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
पाठ 7 Conic Sections
Miscellaneous Exercise 7 | Q 2.2 | पृष्ठ १७८

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