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Three normals are drawn from the point (7, 14) to the parabola `x^2-8x-16 y=0`. Find the coordinates of the feet of the normals. |
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Answer» The equation of the given parabola is `x^(2)-8x-16y=0` Differentiating w.r.t. x, we get `2x-8-16(dy)/(dx)=0or(dy)/(dx)=(x-4)/(8)` Let normal be drawn at point (h,k) on the curve. `:." "k=(h^(2)-8h)/(16)` Also, slope of the normal at `(h,k)=(8)/(4-h)` Therefore, equation of the normal at (h,k) is `y-kj=((8)/(4-h) )(x-h)` If the normal passes through (7,14), then we have `14-(h^(2)-8h)/(16)=((8)/(4-h))(7-h)` `rArr" "(4-h)(224-h^(2)+8h)=128(7-h)` `rArr" "h^(3)-12h^(2)-64h=0` `rArr" "h(h-16)(h+4)=0` `rArr" "h=0,-4,16` For h=0, k=0 for h=-4, k=3 and for h=16, k=8. Therefore, the coordinates of the feet of the normal are (0, 0), (-4,3) and (16,8). |
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