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  • Stuck here, help me understand: - Co-ordinate geometry - JEE Main-6

If a\neq 0 and the line 2bx+3cy+4d=0  passes through the points of intersection of the parabolas y^{2}=4ax\; and\; x^{2}=4ay,then

  • Option 1)

    d^{2}+(2b-3c)^{2}=0

  • Option 2)

    d^{2}+(3b+2c)^{2}=0

  • Option 3)

    d^{2}+(2b+3c)^{2}=0

  • Option 4)

    d^{2}+(3b-2c)^{2}=0

 

Answers (1)

best_answer

As we learnt in

Standard equation of parabola -

y^{2}=4ax

- wherein

 

 

 

Standard equation of parabola -

x^{2}=4ay

- wherein

 

 y^{2}=4ax\: and \: x^{2}=4ay

we get (0,0) and (4a, 4a)

 

Also 2bx+3y+4d=0

Put (0,0) we get d=0

put (4a, 4a) we get 

2b+3c=0

d^{2}+(2b+3c)^{2}=0


Option 1)

d^{2}+(2b-3c)^{2}=0

Incorrect option    

Option 2)

d^{2}+(3b+2c)^{2}=0

Incorrect option    

Option 3)

d^{2}+(2b+3c)^{2}=0

Correct option

Option 4)

d^{2}+(3b-2c)^{2}=0

Incorrect option    

Posted by

Aadil

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