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  • Let the line and the ellipse intersect at a point P in the firs

Let the line y=mx and the ellipse 2x^{2}+y^{2}=1 intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at \left ( -\frac{1}{3\sqrt{2}},0 \right ) and (0,\beta ), then \beta is equal to :
Option: 1 \frac{2}{\sqrt{3}}
Option: 2 \frac{2}{3}
Option: 3 \frac{2\sqrt{2}}{3}
Option: 4 \frac{\sqrt{2}}{3}
 

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}.}

-

\\ {\text { Let } P \text { be }\left(x_{1}, y_{1}\right)} \\\\ {\text { Equation of normal at } P \text { is } \frac{x}{2 x_{1}}-\frac{y}{y_{1}}=-\frac{1}{2}} \\\\ {\text { It passes through }\left(-\frac{1}{3 \sqrt{2}}, 0\right) \Rightarrow \frac{-1}{6 \sqrt{2} x_{1}}=-\frac{1}{2} \Rightarrow x_{1}=\frac{1}{3 \sqrt{2}}}

\\ {\text { So } y_{1}=\frac{2 \sqrt{2}}{3} \text { (as } P \text { lies in } 1 \text { 'quadrant) }} \\\\ {\text { So } \beta=\frac{y_{1}}{2}=\frac{\sqrt{2}}{3}}

Correct option (4)

Posted by

Kuldeep Maurya

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