# 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

#### 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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