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  • Which of the following condition is true if the normal at an end of a latus rectum of an ellipse <img alt="\frac{x^{2}}{a^2}+\frac{y^{2}}{b^2}=1" src="https://learn.careers360.com/latex-image/

Which of the following condition is true if the normal at an end of a latus rectum of an ellipse \frac{x^{2}}{a^2}+\frac{y^{2}}{b^2}=1 passes through one extremity of the minor axis is

Option: 1

e^{4}+e^{2}+1=0


Option: 2

e^{4}+e^{2}-1=0


Option: 3

e^{4}-e^{2}-1=0


Option: 4

e^{4}-e^{2}+1=0


Answers (1)

best_answer

 

 

Equation of Normal in Point Form and Parametric Form -

Equation of Normal in Point Form and Parametric Form

 

Point form:
\\ {\text {The equation of normal at }\left(x_{1}, y_{1}\right) \text { to the ellipse, } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { is }} \\ {\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2}-b^{2}.}

-

 

 

Let one end of a latus rectum is L\left(a e, \frac{b^{2}}{a}\right) and minor axis be B(0,-b)

The equation of the normal to the ellipse at L is

\frac{a}{e} x-a y=a^{2}-b^{2}

which is passing through B^{\prime}(0,-b) so we have

\begin{aligned} & a b=a^{2}-b^{2} \\ \Rightarrow & a b=a^{2}-a^{2}(1-32)=a^{2} e^{2} \\ \Rightarrow & b=a e^{2} \\ \Rightarrow & b^{2}=a^{2} c^{4} \\ \Rightarrow & a^{2}\left(1-e^{2}\right)=a^{2} e^{4} \\ \Rightarrow & e^{4}+e^{2}-1=0 \end{aligned}

 

 

Posted by

vishal kumar

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