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Parabola question

2006 · Shift 0 · Q80
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Parabola question

2006 · Shift 0 · Q80

JEE MainMathematicsParabolaMCQ+4 / −1
The locus of the vertices of the family of parabolas y=a3x23+a2x2−2ay = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2ay=3a3x2​+2a2x​−2a is :
  1. A
    xy=10564xy = {{105} \over {64}}xy=64105​
  2. B
    xy=34xy = {{3} \over {4}}xy=43​
  3. C
    xy=3516xy = {{35} \over {16}}xy=1635​
  4. D
    xy=64105xy = {{64} \over {105}}xy=10564​
View written solutionFree

Correct answer: A

  1. Given family of parabolas

We have y=a3x23+a2x2−2ay=\frac{a^3x^2}{3}+\frac{a^2x}{2}-2ay=3a3x2​+2a2x​−2a where aaa is the parameter.

This is a parabola of the form y=Ax2+Bx+Cy=Ax^2+Bx+Cy=Ax2+Bx+C with A=a33,B=a22,C=−2a.A=\frac{a^3}{3},\qquad B=\frac{a^2}{2},\qquad C=-2a.A=3a3​,B=2a2​,C=−2a.

  1. Vertex of a parabola

For a parabola y=Ax2+Bx+C,y=Ax^2+Bx+C,y=Ax2+Bx+C, the xxx-coordinate of the vertex is xv=−B2A.x_v=-\frac{B}{2A}.xv​=−2AB​.

So here,

Now simplify:

=-\frac{3}{4a}.$$ --- 3. **Find the corresponding $y$-coordinate** Substitute $x=-\dfrac{3}{4a}$ into $$y=\frac{a^3x^2}{3}+\frac{a^2x}{2}-2a.$$ First, $$x^2=\frac{9}{16a^2}.$$ Therefore, $$\frac{a^3x^2}{3}=\frac{a^3}{3}\cdot \frac{9}{16a^2}=\frac{3a}{16}.$$ Next, $$\frac{a^2x}{2}=\frac{a^2}{2}\cdot \left(-\frac{3}{4a}\right)=-\frac{3a}{8}.$$ Hence, $$y_v=\frac{3a}{16}-\frac{3a}{8}-2a.$$ Taking LCM $16$, $$y_v=\frac{3a-6a-32a}{16}=-\frac{35a}{16}.$$ So the vertex is $$\left(-\frac{3}{4a},-\frac{35a}{16}\right).$$ --- 4. **Eliminate the parameter $a$** Let the vertex be $(x,y)$. Then $$x=-\frac{3}{4a},\qquad y=-\frac{35a}{16}.$$ Multiply: $$xy=\left(-\frac{3}{4a}\right)\left(-\frac{35a}{16}\right) =\frac{105}{64}.$$ Thus the locus is $$\boxed{xy=\frac{105}{64}}.$$ --- 5. **Check with options** - A: $xy=\dfrac{105}{64}$ ✓ - B: $xy=\dfrac{3}{4}$ - C: $xy=\dfrac{35}{16}$ - D: $xy=\dfrac{64}{105}$ So the correct option is **A**.
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