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  • Let   <img alt="\mathrm {A(0,3), B\: \: \& \: \: C}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7BA%280%2C3%29%2C%20B%5C%3A%20%5C%3A%20%5C%26%20%5C%3A%20%5C%3A%

Let   \mathrm {A(0,3), B\: \: \& \: \: C}  are points on the parabola \mathrm { y^2=x+9 }  such that, after joining the points   \mathrm { A, B }  and \mathrm { C } , we get a right angled triangle right angled at  \mathrm { B }. Range of ordinate of \mathrm { C } is
 

Option: 1

(-\infty, 1) \cup(5, \infty)


Option: 2

(-\infty, 1] \cup[5, \infty)


Option: 3

(-\infty, 1) \cup[5, \infty)


Option: 4

(1,5)


Answers (1)

best_answer

Given parabola is   \mathrm {y^2=x+9, }  let the ordinate of B and C are  t1 & t respectively.


 \mathrm { \therefore \quad A(0,3), B\left(t_1^2-9, t_1\right), C\left(t^2-9, t\right) }


Now \mathrm { \triangle A B C }  is right angle triangle at \mathrm { B } \mathrm { \therefore A B \perp B C }


{ \begin{aligned} & \mathrm {\text { Now, } \frac{3-t_1}{9-t_1^2} \cdot \frac{t_1-t}{t_1^2-t^2}=-1} \\ & \mathrm {\Rightarrow \frac{1}{3+t_1} \cdot \frac{1}{t_1+t}=-1 \Rightarrow t_1{ }^2+(3+t) t_1+(3 t+1)=0} \end{aligned} }

For real value of  t_1 , we have  (D \geq 0) 


\begin{aligned} &\mathrm { (3+t)^2-4(3 t+1) \geq 0 }\\ \Rightarrow & \mathrm {t^2-6 t+5 \geq 0 \Rightarrow(t-1)(t-5) \geq 0} \end{aligned}

\begin{aligned} &\mathrm { \Rightarrow t \leq 1 \text { or } t \geq 5 \text { (using wavy curve method) } }\\ & \mathrm {\therefore t \in(-\infty, 1] \cup[5, \infty)} \end{aligned}

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Pankaj

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