InterviewSolution
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InterviewSolution
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If `a!=0` and the line `2bx+3cy+4d=0` passes through the points of intersection of the parabola `y^2 = 4ax` and `x^2 = 4ay`, then |
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Answer» Intersection points of `y^2 = 4ax` and `x^2 = 4ay` will be `(0,0)` and `(4a,4a)`. Putting these values of `(x,y)` in given line, `2bx+3cy+4d = 0` At point `(0,0),` `0+0+4d = 0 => d = 0->(1)` At point `(4a,4a),` ` 8ab+12ac+4ad = 0` `=>4a(2b+3c+0) = 0` `=>4a(2b+3c) = 0` As ,` a !=0,` So, `2b+3c = 0->(2)` Squaring and adding (1) and (2), `d^2+(2b+3c)^2 = 0` So, option `D` is the correct option. |
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