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प्रश्न
Find the equations of the tangent and normal to the curve `x^2/a^2−y^2/b^2=1` at the point `(sqrt2a,b)` .
उत्तर
The equation of the given curve is
`x^2/a^2−y^2/b^2=1`
Differentiating with respect to x, we get:
`dy/dx=b^2/a^2 x/y`
`(dy/dx)_(sqrt2a,b)=(sqrt2b)/a`
Therefore, the slope of the tangent is `(sqrt2b)/a` and of the normal is `-a/(sqrtb)`
Thus, the equation of the tangent is
`y-b=(sqrt2b)/a(x-sqrt2a)`
`=>sqrt2bx-ay-ab=0`
Equation of the normal is
`y−b=−a/(sqrt2b)(x−sqrt2a)`
`⇒ax+sqrt2by−sqrt2(a^2+b^2)=0`
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