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The locus of point of intersection of tangent to the parabolas y^2=4(x+2)  and  y^2=8(x+2) which are perpendicular to each other is :


 

Option: 1

\begin{aligned} & x+7=0 \\ \end{aligned}


Option: 2

x-y=0 \\


Option: 3

x+3=0 \\


Option: 4

y-x=10


Answers (1)

The equation of any tangent to y^2=4(x+2)_{\text {is }} y=m(x+1)+\frac{1}{m}.......(1)

The equation of any tangent to y^2=8(x+2) \text { is } y=m^{\prime}(x+2)+\frac{2}{m}.......(2)

It is given that (1) and (2) are perpendicular. Therefore,

\begin{aligned} m m^{\prime} & =-1 \\ \Rightarrow m^{\prime} & =\frac{-1}{m} \end{aligned}
Putting m^{\prime}=\frac{-1}{m} ? in (2), we get

y=\left(\frac{-1}{m}\right)(x+2)-2 m......(3)

The point of intersection of (1) and (3) is given by solving (1) and (2)
On subtracting (3) from (1), we get

\begin{aligned} &0=\left(\frac{m+1}{m}\right) x+3\left(\frac{m+1}{m}\right)\\ &\Rightarrow x+3=0 \quad\left[\because \frac{m+1}{m}=0\right] \end{aligned}

Posted by

Ramraj Saini

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