Let the tangent to the curve x^2+2x-4y+9=0 at the point P(1,3) on it meet the y-axis at A.

Question

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

Answer: 292

Accepted Answer

Given, equation of curve is

${x}^{2}+2x-4y+9=0$ ..........(i)

Equation of tangent at $P(1,3)$ to the given curve (i)

$\begin{gathered} x(1)+2(\frac{x+1}{2})-4(\frac{y+3}{2})+9=0 \ \Rightarrow 2x+2x+2-4y-12+18=0 \ \Rightarrow 4x-4y+8=0 \ \Rightarrow x-y+2=0 \end{gathered}$

which is meet the $Y$-axis at $A$

$	herefore A\equiv (0,2)$

Equation of line passing through $P$ and parallel to $x-3y=6$ is $x-3y+8=0$

Since, line (ii) meet the parabola ${y}^{2}=4x$ at $B$

$\begin{gathered} 	herefore {y}^{2}=4(3y-8) \ \Rightarrow {y}^{2}-12y+32=0 \ \Rightarrow (y-4)(y-8)=0 \end{gathered}$

$	herefore $ Possible co-ordinates of $B$ are $(4,4)$ and $(16,8)$.

Since, $(4,4)$ does not satisfies line $2x-3y=8$

Thus, $B$ is $(16,8)$

$	herefore (AB{)}^{2}=(16-0{)}^{2}+(8-2{)}^{2}=256+36=292$

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