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if `y=4x+3` is parallel to a tangent to the parabola `y^2=12x`, then its distance from the normal parallel to the given line isA. `(213)/(sqrt(17))`B. `(219)/(sqrt(17))`C. `(211)/(sqrt(17))`D. `(210)/(sqrt(17))` |
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Answer» Correct Answer - B Given equation of parabola is `y^2=12x` ....(i) On differenting both sides w.r.t. x, we get `2y(dy)/(dx)=12rArr (dy)/(dx)=(6)/(y)` Since , the noraml to the curve is parallel to the line `y=4x+3` `therefore` Slope of normal curve = Slope of line `rArr -(y)/(6)=4rArr y=-24` From Eq.(i),`(-24)^2=12x` `rArr 24xx24xx12x` `rArr x=48` `therefore` Normal point on a curve is `(48-24)`. And distance from `(48,-24)` to the line `4x-y+3=0` is `(4xx48+24+3)/(sqrt(4^2+1^2))=(219)/(sqrt(17))` |
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