The equation of the normal to the curve x216−y29=1 at (8,3√3) is
A
4x−√3y=23
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B
2x+√3y=25
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C
3x−2√3y=6
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D
x+√3y=17
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Solution
The correct option is B2x+√3y=25 Equation of normal in point form is a2xx1+b2yy1=a2+b2 ∴ equation of the normal to the curve x216−y29=1 at (8,3√3) is 16x8+9y3√3=16+9 ⇒2x+√3y=25
Hence, the required equation of the normal is 2x+√3y=25.