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  • Let be a tangent to the parabola <img alt="\mathrm{y^2=4(x+1) }" src="https://entrancecorner.onco

Let \mathrm{L_1} be a tangent to the parabola \mathrm{y^2=4(x+1) }and \mathrm{L_2} be a tangent to the parabola \mathrm{y^2=8(x+2)} such that \mathrm{\mathrm{L}_1 \ and \ \mathrm{L}_2 } intersect at right angles. Then \mathrm{\mathrm{L}_1 \ and \ \mathrm{L}_2 }  meet on the straight line

Option: 1

x+3=0


Option: 2

2 x+1=0


Option: 3

x+2=0


Option: 4

x+2 y=0


Answers (1)

best_answer

\mathrm{\text { We have } y^2=4(x+1)}                   ...[i]

\mathrm{\text { and } y^2=8(x+2)}                            ...[ii]

Tangent to (i) is given by

\mathrm{L_1: t_1 y=(x+1)+t_1^2}               ...[iii]

Tangent to (ii) is given by

\mathrm{L_1: t_2 y=(x+2)+2 t_2^2}               ....[iv]

\mathrm{\text { Now, } L_1 \perp L_2}

\mathrm{\therefore \quad \frac{1}{t_1} \cdot \frac{1}{t_2}=-1 \Rightarrow t_1 t_2=-1}

Multiplying (iii) by \mathrm{t_2} and (iv) by \mathrm{t_1}, and then subtracting, we get

\mathrm{\begin{aligned} & t_2(x+1)-t_1(x+2)+t_1^2 t_2-2 t_1 t_2^2=0 \\ & \Rightarrow \quad\left(t_2-t_1\right) x+t_2-2 t_1-t_1+2 t_2=0 \\ & \Rightarrow \quad\left(t_2-t_1\right) x=-3 t_2+3 t_1=-3\left(t_2-t_1\right) \\ & \Rightarrow \quad x+3=0 \end{aligned}}

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Shailly goel

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