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Parabola question

2020 · 8 Jan · Shift 1 · Q29
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Parabola question

2020 · 8 Jan · Shift 1 · Q29

JEE MainMathematicsParabolaMCQ+4 / −1
The locus of a point which divides the line segment joining the point (0, –1) and a point on the parabola, x2 = 4y, internally in the ratio 1 : 2, is :
  1. A
    9x2 – 3y = 2
  2. B
    4x2 – 3y = 2
  3. C
    x2 – 3y = 2
  4. D
    9x2 – 12y = 8
View written solutionFree

Correct answer: D

  1. Set up the variable point on the parabola

A general point on the parabola x2=4yx^2=4yx2=4y can be written as Q(2t,t2).Q(2t,t^2).Q(2t,t2).

The fixed point is P(0,−1).P(0,-1).P(0,−1).

  1. Use the section formula

Let the required point be R(h,k)R(h,k)R(h,k), which divides the line segment joining P(0,−1)P(0,-1)P(0,−1) and Q(2t,t2)Q(2t,t^2)Q(2t,t2) internally in the ratio 1:21:21:2.

Since RRR divides PQPQPQ internally in the ratio 1:21:21:2, we use the section formula: R(1⋅2t+2⋅01+2,1⋅t2+2⋅(−1)1+2).R\left(\frac{1\cdot 2t+2\cdot 0}{1+2},\frac{1\cdot t^2+2\cdot(-1)}{1+2}\right).R(1+21⋅2t+2⋅0​,1+21⋅t2+2⋅(−1)​).

So, h=2t3,k=t2−23.h=\frac{2t}{3},\qquad k=\frac{t^2-2}{3}.h=32t​,k=3t2−2​.

Thus, x=2t3,y=t2−23.x=\frac{2t}{3},\qquad y=\frac{t^2-2}{3}.x=32t​,y=3t2−2​.

  1. Eliminate the parameter ttt

From x=2t3,x=\frac{2t}{3},x=32t​, we get t=3x2.t=\frac{3x}{2}.t=23x​.

Substitute into the expression for yyy: y=\frac{t^2-2}{3}= rac{\left(\frac{3x}{2}\right)^2-2}{3}.

So, y=\frac{\frac{9x^2}{4}-2}{3}= rac{9x^2-8}{12}.

Multiply by 121212: 12y=9x2−8.12y=9x^2-8.12y=9x2−8.

Hence, 9x2−12y=8.9x^2-12y=8.9x2−12y=8.

  1. Match with the options

The locus is 9x2−12y=8,9x^2-12y=8,9x2−12y=8, which corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So, the derived answer agrees with the stored answer.

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