the normal to ellipse 4x^2+5y^2=20 at the point 'p' touches the parabola y^2=4x,the eccentric angle of the point 'p' is

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Deepak Kumar Shringi · Answer

the answer is =pie-tan -1 (2) im attaching the solution

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the normal to ellipse 4x^2+5y^2=20 at the point 'p' touches the parabola y^2=4x,the eccentric angle of the point 'p' is

the normal to ellipse 4x^2+5y^2=20 at the point 'p' touches the parabola y^2=4x,the eccentric angle of the point 'p' is

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the normal to ellipse 4x^2+5y^2=20 at the point 'p' touches the parabola y^2=4x,the eccentric angle of the point 'p' is

the normal to ellipse 4x^2+5y^2=20 at the point 'p' touches the parabola y^2=4x,the eccentric angle of the point 'p' is

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