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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. |
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Answer» Correct Answer - `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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