Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Tangents are drawn from the point (-1,2) on the parabola The length, these tangen

Tangents are drawn from the point (-1,2) on the parabola \mathrm{y^2=4 x.} The length, these tangents will intercept on the line \mathrm{\mathrm{x}=2} is :

Option: 1

6


Option: 2

6 \sqrt{2}


Option: 3

2 \sqrt{6}


Option: 4

none of these


Answers (1)

best_answer

\mathrm{\begin{aligned} & \mathrm{SS}_1=\mathrm{T}^2 \\ & \left(\mathrm{y}^2-4 \mathrm{x}\right)\left(\mathrm{y}_1{ }^2-4 \mathrm{x}_1\right)=\left(\mathrm{y}_1-2\left(\mathrm{x}+\mathrm{x}_1\right)\right)^2 \\ & \left(\mathrm{y}^2-4 \mathrm{x}\right)(4+4)=[2 \mathrm{y}-2(\mathrm{x}-1)]^2=4(\mathrm{y}-\mathrm{x}+1)^2 \\ & 2\left(\mathrm{y}^2-4 \mathrm{x}\right)=(\mathrm{y}-\mathrm{x}+1)^2 ;\end{aligned}}

solving with the line  \mathrm{\mathrm{x}=2 } we get,

\mathrm{2\left(\mathrm{y}^2-8\right)=(\mathrm{y}-1)^2 \quad \text { or } 2\left(\mathrm{y}^2-8\right)=\mathrm{y}^2-2 \mathrm{y}+1}

or  \mathrm{y^2+2 y-17=0}

\mathrm{\text { where } y_1+y_2=-2 \text { and } y_1 y_2=-17}

\mathrm{\text { Now }\left|\mathrm{y}_1-\mathrm{y}_2\right|^2=\left(\mathrm{y}_1+\mathrm{y}_2\right)^2-4 \mathrm{y}_1 \mathrm{y}_2}

\mathrm{\text { or }\left|y_1-y_2\right|^2=4-4(-17)=72}

\mathrm{\therefore\left(\mathrm{y}_1-\mathrm{y}_2\right)=\sqrt{72}=6 \sqrt{2}}

Posted by

Ritika Harsh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question