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प्रश्न
Find the equation of the tangent to the hyperbola:
9x2 – 16y2 = 144 at the point L of latus rectum in the first quadrant
उत्तर
The equation of the hyperbola is 9x2 – 16y2 = 144
i.e. `x^2/16 - y^2/9` = 1
Comparing with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = 16, b2 = 9
∴ a = 4, b = 3
Eccentricity = e = `sqrt("a"^2 + "b"^2)/"a"`
= `sqrt(16 + 9)/4`
= `5/4`
∴ ae = `4(5/4)` = 5
and `"b"^2/"a" = 9/4`
∴ the end of latus rectum in the first quadrant is
L = `("ae", "b"^2/"a") = (5, 9/4)`
The equation of the tangent to `x^2/"a"^2 - y^2/"b"^2` = 1 at the point (x1, y1) is `("xx"_1)/"a"^2 - ("yy"_1)/"b"^2` = 1
∴ the equation of the tangent to given hyperbola at L is
`(x(5))/16 - (y(9/4))/9` = 1
∴ `(5x)/16 - y/4` = 1 i.e., 5x – 4y = 16.
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