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Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is :

  • Option 1)

    4\left ( x+y \right )+3=0

     

     

     

  • Option 2)

    3\left ( x+y \right )+4=0

  • Option 3)

    8\left ( 2x+y \right )+3=0

  • Option 4)

    x+2y+3=0

 

Answers (1)

best_answer

As we learned

 

Length of the latus rectum -

4a

 

- wherein

For the parabola.

y^{2}=4ax

 

 

 

Equation of tangent -

y= mx+\frac{a}{m}

- wherein

Tengent to y^{2}=4ax is slope form.

 

 

Common tangent of  x^{2}=3y\; and\; y^{2}=3x

y=mx-\frac{3}{4}m^{2}\; \; \; \; \; \; \; \; \; \; y=mx+\frac{3}{4m}

On solving   \frac{3}{4}m^{2}=\frac{3}{4}m=1\; \; \; \; m=-1

We get 

x+y+\frac{3}{4}=0\Rightarrow 4\left ( x+y \right )+3=0

 

 


Option 1)

4\left ( x+y \right )+3=0

 

 

 

Option 2)

3\left ( x+y \right )+4=0

Option 3)

8\left ( 2x+y \right )+3=0

Option 4)

x+2y+3=0

Posted by

Himanshu

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