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A point on the ellipse `x^2+3y^2=37`where the normal is parallel to the line `6x-5y=2`is`(5,-2)`(b) (5, 2)(c) `(-5,2)`(d) `(-5,-2)`A. (5,-2)B. (5,2)C. (-5,2)D. (-5,-2) |
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Answer» Differnetiating the equation of ellipse, `x^(2)+3y^(2)=37` w.r.t., x we get `(dy)/(dx)=-(x)/(3y)` The slop of the given line is 6/5 , which is normal to the ellipse Hence, `3x//y=6//5 or 2x=5y` Points in options (2) and (4) are satisfying the above equation adn that of ellipse. |
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