If a normal to the parabola y^2=12x at A(3,-6 | vedclass

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(A) The equation of the parabola is $y^2=12x$. Comparing with $y^2=4ax$,we get $a=3$. The slope of the tangent at any point $(x_1, y_1)$ is given by $

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$x-3y+27=0$

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(A) The equation of the parabola is $y^2=12x$. Comparing with $y^2=4ax$,we get $a=3$. The slope of the tangent at any point $(x_1, y_1)$ is given by $\frac{dy}{dx} = \frac{2a}{y_1} = \frac{6}{y_1}$. At $A(3,-6)$,the slope of the tangent is $\frac{6}{-6} = -1$. Therefore,the slope of the normal at $A(3,-6)$ is $m = -1/(-1) = 1$. The equation of the normal at $A(3,-6)$ is $y - (-6) = 1(x - 3)$,which simplifies to $y = x - 9$. To find the intersection point $P$,substitute $y = x - 9$ into $y^2 = 12x$: $(x-9)^2 = 12x$ $\Rightarrow x^2 - 18x + 81 = 12x$ $\Rightarrow x^2 - 30x + 81 = 0$. $(x-3)(x-27) = 0$. Since $x=3$ corresponds to point $A$,the point $P$ has $x=27$. Then $y = 27 - 9 = 18$. So,$P$ is $(27, 18)$. The equation of the tangent at $P(27, 18)$ to $y^2=12x$ is $yy_1 = 2a(x+x_1)$. $y(18) = 2(3)(x+27)$ $\Rightarrow 18y = 6(x+27)$ $\Rightarrow 3y = x+27$ $\Rightarrow x-3y+27=0$.

If a normal to the parabola $y^2=12x$ at $A(3,-6)$ cuts the parabola again at $P$,then the equation of the tangent at $P$ is

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