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प्रश्न
The normal to the hyperbola `x^2/a^2 - y^2/9` = 1 at the point `(8, 3sqrt(3))` on it passes through the point ______.
पर्याय
`(15, -2sqrt(3))`
`(9, 2sqrt(3))`
`(-1, 9sqrt(3))`
`(-1, 6sqrt(3))`
MCQ
रिकाम्या जागा भरा
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उत्तर
The normal to the hyperbola `x^2/a^2 - y^2/9` = 1 at the point `(8, 3sqrt(3))` on it passes through the point `underlinebb((-1, 9sqrt(3))`.
Explanation:
Given hyperbola is `x^2/a^2 - y^2/9` = 1
∵ `(8, 3sqrt(3))` lie on hyperbola then point will satisfy it
`\implies 64/a^2 - 27/9` = 1
`\implies a^2 = 64/4` = 16
Now, equation of normal at `(8, 3sqrt(3))` :
`(16x)/8 + (9y)/(3sqrt(3))` = 16 + 9
`2x + sqrt(3)y` = 25
On putting x, y from the given options, we get option (c) is correct.
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Application of Derivative in Geometry
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