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Find the equation of the tangent to the hyperbola: 3x2 – 4y2 = 12 at the point (4, 3) - Mathematics and Statistics

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प्रश्न

Find the equation of the tangent to the hyperbola:

3x2 – 4y2 = 12 at the point (4, 3)

बेरीज
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उत्तर

The equation of the hyperbola is 3x2 – 4y2 = 12, i.e., `x^2/4 - y^2/3` = 1

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

a2 = 4, b2 = 3

The equation of the tangent to the hyperbola

`x^2/"a"^2 - y^2/"b"^2` = 1 at the point P(x1, y1) on it is

`("xx"_1)/"a"^2 - ("yy"_1)/"b"^2` = 1

∴ the equation of the tangent to the given hyperbola at the point (4, 3) is

`("x"(4))/4 - ("y"(3))/3 = 1`

∴ x – y = 1

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Conic Sections - Hyperbola
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Conic Sections - Exercise 7.3 [पृष्ठ १७५]

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