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Find the equations of the tangent and normal to the curve x2/a2−y2/b2=1 at the point (√2a,b) - Mathematics

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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)` .

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Solution

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`

Concept: Tangents and Normals
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