Q&A - Ask Doubts and Get Answers
Q

Solve! - Co-ordinate geometry - JEE Main-2

The locus of the vertices of the family of parabolas y=\frac{a^{3}x^{2}}{3}+\frac{a^{2}x}{2}-2a\; \; is

  • Option 1)

    xy=\frac{105}{64}

  • Option 2)

    xy=\frac{3}{4}

  • Option 3)

    xy=\frac{35}{16}

  • Option 4)

    xy=\frac{64}{105}

 
Answers (1)
141 Views
V Vakul

As we learnt in 

Rectangular Hyperbola -

x^{2}-y^{2}= a^{2}

- wherein

 

 and

Standard equation of parabola -

y^{2}=4ax

- wherein

 

 Family of parabolas

y=\frac{a^{2}x^{2}}{3} + \frac{a^{2}x}{2}-2a

For quadratic, A2x2+Bx + C

vertex is \left(\frac{-B}{2A}, \frac{-D}{4A} \right )

so, h=\frac{-a^{2}/2}{2\times a^{2}/3}=\frac{-3}{4a}

K=\frac{\left ( \frac{a^{2}}{2} \right )^{2}-4a^{3}\times \left ( \frac{-2a}{3} \right )}{4\left ( \frac{a^{3}}{3} \right )}=-\frac{35}{16}a

Eliminating a from h and k

hk=-\frac{3}{4a}\times\frac{-35}{16}a=\frac{105}{64}

xy=\frac{105}{64}


Option 1)

xy=\frac{105}{64}

This is correct option

Option 2)

xy=\frac{3}{4}

This is incorrect option

Option 3)

xy=\frac{35}{16}

This is incorrect option

Option 4)

xy=\frac{64}{105}

This is incorrect option

Exams
Articles
Questions