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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 parabola is
 

Option: 1

\mathrm{4(x+y)+3=0}
 


Option: 2

\mathrm{8(2 x+y)+3=0}
 


Option: 3

\mathrm{3(x+y)+4=0}
 


Option: 4

\mathrm{x+2 y+3=0}


Answers (1)

\mathrm{\because \text{Length of latus rectum of each parabola} =3}

\mathrm{\therefore \quad \text{Equation of parabolas be}\: y^2=3 x}      .......(i)

\mathrm{and \: \: x^2=3 y}          ..........(ii)

The equation of tangent to (i) is

\mathrm{ y=m x+\frac{3}{4 m} }...........(iii)

(iii) is also tangent to \mathrm{ x^2=3 y }

\mathrm{ \Rightarrow x^2=3 m x+\frac{9}{4 m} }

\mathrm{ \Rightarrow 4 m x^2=12 m^2 x+9 \Rightarrow 4 m x^2-12 m^2 x-9=0 }

\mathrm{ Now, D=0 \Rightarrow 144 m^4-4(-9)(4 m)=0 }

\mathrm{ \Rightarrow m\left(m^3+1\right)=0 \Rightarrow m=0 (\text{which is not possible}) }

\mathrm{ \therefore m=-1 }

\mathrm{ Put \: m=-1 \: in (iii), we\: \: get\: y=-x-\frac{3}{4} }

\mathrm{ or \: 4(x+y)+3=0, \text{which is the required equation of tangent.} }

Hence option 1 is correct.





 

Posted by

Ramraj Saini

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