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Find the maximum distance of any normal ...

Find the maximum distance of any normal of the ellipse `x^2/a^2 + y^2/b^2=1` from its centre,

Text Solution

Verified by Experts

Normally to ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at point `P(a cos theta, b sin theta)` is `(ax)/(cos theta)-(by)/(sin theta)=a^(2)-b^(2)`
Its distance from origin is
`d=(|a^(2)-b^(2)|)/(sqrt(a^(2)sec^(2)theta+b^(2)"cosec"^(2)theta))`
`d=(|a^(2)-b^(2)|)/(sqrt(a^(2)+b^(2)+a^(2)tan^(2)theta+b^(2)cot^(2)theta))`
`d=(|a^(2)-b^(2)|)/(sqrt(a^(2)+b^(2)+2ab+(atan theta-bcottheta)^(2)))`
`d=(|a^(2)-b^(2)|)/(sqrt(a^(2)+b^(2)+2ab))`
`:. d_("max")=a-b)`
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