The focus of the parabola y^2 = 4x + 16 is th | vedclass

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(C) The given parabola is $y^2 = 4(x + 4)$. Comparing this with $y^2 = 4a(x - h)$,we get $a = 1$ and the vertex is $(-4, 0)$. The focus is $(h+a, k) =

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$15$

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(C) The given parabola is $y^2 = 4(x + 4)$. Comparing this with $y^2 = 4a(x - h)$,we get $a = 1$ and the vertex is $(-4, 0)$. The focus is $(h+a, k) = (-4+1, 0) = (-3, 0)$.Since the focus is the centre of the circle $C$ with radius $r = 5$,the equation of the circle is $(x + 3)^2 + y^2 = 5^2 = 25$.The point of intersection of the lines $3x - y = 0$ (or $y = 3x$) and $x + \lambda y = 4$ is found by substituting $y = 3x$ into the second equation: $x + \lambda(3x) = 4 \implies x(1 + 3\lambda) = 4 \implies x = \frac{4}{1 + 3\lambda}$. Then $y = \frac{12}{1 + 3\lambda}$.Since the circle passes through this point,we substitute the coordinates into the circle equation: $(\frac{4}{1 + 3\lambda} + 3)^2 + (\frac{12}{1 + 3\lambda})^2 = 25$.Simplifying: $(\frac{4 + 3 + 9\lambda}{1 + 3\lambda})^2 + \frac{144}{(1 + 3\lambda)^2} = 25 \implies (7 + 9\lambda)^2 + 144 = 25(1 + 3\lambda)^2$.$49 + 126\lambda + 81\lambda^2 + 144 = 25(1 + 6\lambda + 9\lambda^2) \implies 81\lambda^2 + 126\lambda + 193 = 225\lambda^2 + 150\lambda + 25$.$144\lambda^2 + 24\lambda - 168 = 0 \implies 6\lambda^2 + \lambda - 7 = 0$.$(6\lambda + 7)(\lambda - 1) = 0$,so $\lambda_1 = -7/6$ and $\lambda_2 = 1$.Finally,$12\lambda_1 + 29\lambda_2 = 12(-7/6) + 29(1) = -14 + 29 = 15$.

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