Let the tangent to the curve x^2+2x-4y+9=0 at | vedclass

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(D) The equation of the curve is $x^2+2x-4y+9=0$. The tangent at $P(1,3)$ is given by $x(1) + (x+1) - 2(y+3) + 9 = 0$,which simplifies to $x+1+x-2y-6+

Vedclass · Accepted Answer

$292$

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(D) The equation of the curve is $x^2+2x-4y+9=0$. The tangent at $P(1,3)$ is given by $x(1) + (x+1) - 2(y+3) + 9 = 0$,which simplifies to $x+1+x-2y-6+9=0$,or $2x-2y+4=0$,which is $x-y+2=0$. Setting $x=0$ for the $y$-axis,we get $y=2$,so $A = (0,2)$. The line passing through $P(1,3)$ parallel to $x-3y=6$ has the equation $x-3y = 1-3(3) = -8$,or $x-3y+8=0$. This line meets $y^2=4x$. Substituting $x=3y-8$ into $y^2=4x$,we get $y^2=4(3y-8) \implies y^2-12y+32=0$. Factoring gives $(y-4)(y-8)=0$,so $y=4$ or $y=8$. If $y=4$,$x=3(4)-8=4$. If $y=8$,$x=3(8)-8=16$. The points are $(4,4)$ and $(16,8)$. Checking the condition $2x-3y=8$: For $(4,4)$,$2(4)-3(4) = 8-12 = -4 
eq 8$. For $(16,8)$,$2(16)-3(8) = 32-24 = 8$. Thus,$B = (16,8)$. Finally,$(AB)^2 = (16-0)^2 + (8-2)^2 = 16^2 + 6^2 = 256 + 36 = 292$.

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 $............$.

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