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If x/a+y/b=sqrt(2) touches the ellipse (...

If `x/a+y/b=sqrt(2)` touches the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` , then find the eccentric angle `theta` of point of contact.

Text Solution

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Equation of tangent to ellipse at`P( a cos theta, b sin theta )` is `(x cos theta)/(a)+(y sin theta)/(b)=1`
Also, `(x)/(asqrt(2))+(y)/(bsqrt(2))=1` is the tangent. Therefore,
`(cos theta)/(1//sqrt(2))+(sin theta)/(1//sqrt(2))=1`
`or cos theta=(1)/(sqrt(2)),sin theta=(1)/(sqrt(2))`
`or theta= 45^(@)`
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