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  • The equation of the normal to the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\

The equation of the normal to the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} at the positive end of the latus-rectum is

Option: 1

\mathrm{x+e y+e^3 a=0}


Option: 2

\mathrm{x-e y-e^3 a=0}


Option: 3

\mathrm{x-e y-e^2 a=0}


Option: 4

None of these


Answers (1)

best_answer

The equation of the normal at \mathrm{\left(x_1, y_1\right)} to the given ellipse is \mathrm{\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2} \mathrm{\text { Here, } \quad x_1=a e} 
\mathrm{\text { and } y_1=\frac{b^2}{a}} 
So, the equation of the normal at positive end of the latus- rectum is
\mathrm{\frac{a^2 x}{a e}-\frac{b^2 y}{b^2 / a}=a^2 e^2} \mathrm{\left[\begin{array}{ll} \Delta & b^2=a^2\left(1-e^2\right) \end{array}\right] \Rightarrow} \mathrm{\frac{a x}{e}-a y=a^2 e^2 \Rightarrow x-e y-e^3 a=0}

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Deependra Verma

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