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The slope of normal to be parabola `y = (x^(2))/(4) -2` drawn through the point `(10,-1)` isA. `-2`B. `-sqrt(3)`C. `-1//2`D. `-5//3` |
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Answer» Correct Answer - C `x^(2) = 4(y+2)` is the given parabola. Any normal is `x = m (y+2) -2m -m^(3)` If `(10,-1)` lies on this line then `10 = m -2m -m^(3)` `rArrm^(3) + m + 10 =0 rArr m =-2` Slope of normal `= 1//m =- 1//2`. |
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