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The point on the ellipse x^(2)+2y^(2)=6which is nearest to the line x-y=7 is |
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Answer» `((sqrt6)/(sqrt5),(SQRT3)/(sqrt5))` Normal at P is perpendicular to `x-y=7` `therefore""1=-(sqrt6 cos theta)/(2sqrt3 sin theta), tan theta=(1)/(SQRT2), theta" line in "4^("th") " QUADRANT, "sin theta=-(1)/(sqrt3), cos theta=(sqrt2)/(sqrt3)` `P(2, -1)` |
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