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Find the equation of the normal to the e...

Find the equation of the normal to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` at the positive end of the latus rectum.

Text Solution

Verified by Experts

The correct Answer is:
`x-ey-e^(3)a=0`
The equation of the normal at `(x_(1),y_(2))`, to the given ellipse is
`(a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2)`
Here, `x_(1)` =ad and `y_(1)=b^(2)//a`
So, the equatio of the normal at the positive end of the latus rectum is
`(a^(2)x)/(ae)-(b^(2)y)/(b^(2)//a)=a^(2)e^(2)" "[ :. b^(2)=a^(2)(1-e^(2))]`
or `(ax)/(e)-ay=^(2)e^(2)or x-ey-e^(3)a=0`
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