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  • Find the equation of the common tangents to the parabolas and <img alt="x= -(y- 2)^2" src="https://entranceco

Find the equation of the common tangents to the parabolas x =y^2 and x= -(y- 2)^2.

Option: 1

4x+y+4=0


Option: 2

4x-y+4=0


Option: 3

x-4y+4=0


Option: 4

x+4y+4=0


Answers (1)

best_answer

 

 

Tangents of Parabola in Parametric Form -

Tangents of Parabola in Parametric Form

\begin{array}{c||c cl} \\\mathbf { {Equation \;of \;Parabola} } & {\mathbf { Coordinate }} & {\mathbf { Tangent\; Equation }}\\ \\ \hline\hline\\ {\color{Teal} y^{2}=4ax} & {\color{Teal} {\left(at^{2}, 2 a t\right)}} & {\color{Teal} {t y=x+a t^{2}}} \\ \\ {\color{Red} x^{2} {=4 a y}} & {\color{Red} {(2 a t, a t^2)}} & {\color{Red} {t x=y+a t^{2}}} \\ \\ {\color{Teal} y^{2}{=-4 a x}} & {\color{Teal} {\left(-a t^{2}, 2 a t\right)} }& {\color{Teal} {t y=-x+a t^{2}}} \\ \\ {\color{Red} x^{2} {=} {-4 a y}} & {\color{Red} {\left(2 a t,-at^{2}\right)}} & {\color{Red} {t x=-y+a t^{2}} }\\ \end{array}

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Any point on the parabola x=y^{2} \text { is }\left(t^{2}, t\right)

Now the tangent at \left(t^{2}, t\right) is

\begin{aligned} yy_{1} &=\frac{1}{2}\left(x+x_{1}\right) \\ \Rightarrow \quad t y &=\frac{1}{2}\left(x+t^{2}\right) \\ \end{aligned}\\\Rightarrow \quad x- 2 t y+t^{2} =0

If it is a tangent to the parabola,

x =-(y-2)^{2}, \text { then } \\ {2 t y-t^{2} =-(y-2)^{2}} \\ { \Rightarrow 2 t y-t^{2} =-y^{2}+4 y-4} \\ \Rightarrow y^{2}-2(2-t) y+\left(4-t^{2}\right)=0

Since it has equal roots,

D =0 \\ 4(2-t)^{2}-4\left(t^{2}-4\right) =0 \\ \Rightarrow \quad(2-t)^{2}-\left(t^{2}-4\right)=0 \\ \Rightarrow \quad t=2

Hence, the equation of the common tangent is

x-4y+4=0

 

 

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