Let L_1, L_2 be the lines passing through the | vedclass

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(A) The equation of the parabola is $9x^2+12x+18y-14=0$.Rewriting it,we get $9x^2+12x+4 = -18y+14+4$,which simplifies to $(3x+2)^2 = -18(y-1)$.Let the

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$68$

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(A) The equation of the parabola is $9x^2+12x+18y-14=0$.Rewriting it,we get $9x^2+12x+4 = -18y+14+4$,which simplifies to $(3x+2)^2 = -18(y-1)$.Let the lines passing through $P(0,1)$ be $y-1 = mx$,or $y = mx+1$.Substituting this into the parabola equation: $(3x+2)^2 = -18(mx+1-1) = -18mx$.$9x^2+12x+4 = -18mx \implies 9x^2+(12+18m)x+4 = 0$.Since the lines are tangent,the discriminant $D = 0$.$(12+18m)^2 - 4(9)(4) = 0 \implies (12+18m)^2 = 144$.$12+18m = 12$ or $12+18m = -12$.$18m = 0 \implies m_1 = 0$ or $18m = -24 \implies m_2 = -4/3$.Let $	heta$ be the angle between the tangents. $	an 	heta = |(m_1-m_2)/(1+m_1m_2)| = |(0 - (-4/3))/(1+0)| = 4/3$.For $	riangle PQR$ to be isosceles with base $QR$,the line $QR$ must be perpendicular to the angle bisector of the angle between the tangents.The angle of the bisector is $\alpha = 	heta/2 + 90^\circ$ (or similar geometry). The slope of $QR$ is $m = -\cot(	heta/2)$.Using $	an 	heta = 4/3$,we have $2	an(	heta/2)/(1-	an^2(	heta/2)) = 4/3$.$3	an(	heta/2) = 2 - 2	an^2(	heta/2) \implies 2	an^2(	heta/2) + 3	an(	heta/2) - 2 = 0$.$(2	an(	heta/2)-1)(	an(	heta/2)+2) = 0$.So $	an(	heta/2) = 1/2$ or $-2$.The slopes are $m = -\cot(	heta/2)$,so $m_1 = -2$ and $m_2 = 1/2$.Then $16(m_1^2+m_2^2) = 16(4 + 1/4) = 16(17/4) = 68$.

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