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  • From any point <img alt="\mathrm{P\left(\mathrm{at}^2\right., 2at) }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BP%5Cleft%28%5Cmathrm%7Bat%7D%5E2%5Cright.%2C%202at%29%20%7D" /

From any point \mathrm{P\left(\mathrm{at}^2\right., 2at) }on the parabola \mathrm{\mathrm{y}^2=4 \mathrm{ax}}, two normals can be drawn and their feet \mathrm{Q} and \mathrm{ R} have the parameters satisfying the equation \mathrm{ \lambda^2+\lambda t+k=0}, where \mathrm{ k=}

Option: 1

1


Option: 2

-1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

\mathrm{ \begin{aligned} & \mathrm{t}=-\lambda_1-\frac{2}{\lambda_1}=-\lambda_2-\frac{2}{\lambda_2} \\ & \Rightarrow\left(\lambda_2-\lambda_1\right)=2\left(\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right) \\ & \Rightarrow \lambda_1 \lambda_2=2 \text { and } 2 \mathrm{a} \lambda_1+2 \mathrm{a} \lambda_2+2 \mathrm{at}=0 \\ & \Rightarrow \lambda_1+\lambda_2=-\mathrm{t} \end{aligned} }

\Rightarrow Equation \mathrm{ \lambda^2+\lambda t+2=0}, the roots of

    parameters are feet of Q and R.

Posted by

Divya Prakash Singh

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