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  • Let the tangents to the curve at the point <img alt="P(1,3)" sr

Let the tangents to the curve x^{2}+2 x-4 y+9=0 at the point P(1,3) on it meet the y-axis at A. Let the line passing through \mathrm{P} and parallel to the line \mathrm{x}-3 \mathrm{y}=6meet the parabola \mathrm{y}^{2}=4 \mathrm{x} at \mathrm{B}. If \mathrm{B} lies on the line 2 \mathrm{x}-3 y=8 , then (A B)^{2} is equal to ____.

Option: 1

292


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

C: \mathrm{x}^{2}+2 \mathrm{x}-4 \mathrm{y}+9=0
C:(x+1)^{2}=4(\mathrm{y}-2)

\mathrm{T}_{\mathrm{P}(1,3)}: \mathrm{x} .1+(\mathrm{x}+1)-2(\mathrm{y}+3)+9=0
2 \mathrm{x}-2 \mathrm{y}+4=0
\mathrm{T}_{\mathrm{p}}: \mathrm{x}-\mathrm{y}+2=0
A : (0,2)

Line \|$ to $x-3 y=6$ passes $(1,3)$ is $x-3 y+8=0
Meet\: parabola:y^{2}= 4x

\Rightarrow \mathrm{y}^{2}=4(3 \mathrm{y}-8)
\Rightarrow \mathrm{y}^{2}-12 \mathrm{y}+32=0
\Rightarrow \mathrm{y}^{2}-12 \mathrm{y}+32=0
\Rightarrow(\mathrm{y}-8)(\mathrm{y}-4)=0
\Rightarrowpoint of intersection are

(4,4) \&(16,8)$ lies on $2 x-3 y=8

Hence \; A : (0,2)
\mathrm{B}:(16,8)
(A B)^{2}=256+36=292

Posted by

Deependra Verma

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