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  • Two straight lines are perpendicular to each other. One of them touches the parabola y2 = 4a(x + a) and the other touches x2 = 4b(y + b).The point of intersection of

Two straight lines are perpendicular to each other. One of them touches the parabola y2 = 4a(x + a) and the other touches x2 = 4b(y + b).The point of intersection of the lines is

Option: 1

y+a+b=0


Option: 2

x+a+b=0


Option: 3

y=a+b


Option: 4

x=a+b


Answers (1)

best_answer

 

 

Point of Intersection of Tangent -

Point of Intersection of Tangent

\\ {\text { Two points, } P \equiv\left(a t_{1}^{2}, 2 a t_{1}\right) \text { and } Q \equiv\left(a t_{2}^{2}, 2 a t\right) \text { on the parabola } y^{2}=4 a x .} \\ {\text { Then, equation of tangents at } P \text { and } Q \text { are }} \\ {t_{1} y=x+a t_{1}^{2}} \\ {t_{2} y=x+a t_{1}^{2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots() \\ {\text { solving (i) at }_{2}^{2}} \\ {\text { we get, } x=a t_{1} t_{2}, y=a\left(t_{1}+t_{2}\right)} \\ {\text { Point of Intersection of tangents drawn at point } P \text { and } Q \text { is }} \\ {\left(a t_{1} t_{2}, a\left(t_{1}+t_{2}\right)\right)}

 

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\begin{array}{l}{\text { Any tangent to } y^{2}=4(x+a) \text { is }} \\\\ {\qquad y=m_{1}(x+a)+\frac{a}{m_{1}}}\quad.......(i) \\\\ {\text { Also, any tangent to } y^{2}=4 b(x+b) \text { is }} \\\\ {\qquad y=m_{2}(x+b)+\frac{b}{m_{2}}} \quad.......(ii)\\\\ {\text { since, two tangents are perpendicular, so }} \\ {\qquad m_{1} m_{2}=-1} \\\\ {\Rightarrow \quad m_{2}=-\frac{1}{m_{1}}}\end{array}

From Eq. (ii), we get,

\begin{array}{l}{\text { }} \\ {\qquad y=-\frac{1}{m_{1}}(x+b)-b m_{1}}\quad.......(iii) \\\\ {\text { Now subtracting Eq. (i) and Eq. (iii), we get }} \\\\ {\qquad m_{1}(x+a)+\frac{1}{m_{1}}(x+b)+\frac{a}{m_{1}}+b m_{1}=0} \\\\ {\Rightarrow \quad\left(m_{1}+\frac{1}{m_{1}}\right) x+\left(m_{1}+\frac{1}{m_{1}}\right) a+\left(m_{1}+\frac{1}{m_{1}}\right) b=0} \\\\ {\Rightarrow \quad x+a+b=0}\end{array}

 

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seema garhwal

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