Maharashtra State BoardHSC Science (Electronics) 11th
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Find the equation of the tangent to the hyperbola: x216-y29 = 1 at the point in a first quadratures whose ordinate is 3 - Mathematics and Statistics

Sum

Find the equation of the tangent to the hyperbola:

`x^2/16 - y^2/9` = 1 at the point in a first quadratures whose ordinate is 3

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Solution

Given equation of the hyperbola is `x^2/16 - y^2/9` = 1.

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

a2 = 16 and b2 = 9

Let P(x1, 3) be the point on the hyperbola in the first quadrant at which the tangent is drawn.

Substituting x = x1 and y = 3 in equation of hyperbola, we get

`x_1^2/16 - 3^2/9` = 1

∴ `x_1^2/16 - 1` = 1

∴ `x_1^2/16` = 2

∴ `x_1^2` = 32

∴ x1 = `±  4sqrt(2)`

Since P lies in the first quadrant,

P ≡ `(4sqrt(2), 3)`

Equation of the tangent to the hyperbola

`x^2/"a"^2 - y^2/"b"^2` = 1 at (x1, y1) is

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

Equation of the tangent at P`(4sqrt(2),3)` is

`(4sqrt(2)x)/16 - (3y)/9` = 1

∴ `sqrt(2)/4x - y/3` = 1

∴ `3sqrt(2)x - 4y` = 12

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