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(D) The focal distance of a point $(x_1, y_1)$ on the parabola $y^2=4ax$ is $x_1+a$. Given parabola is $y^2=kx$,so $4a=k \Rightarrow a=k/4$. The focal

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$x \pm \sqrt{2} y+2=0$

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(D) The focal distance of a point $(x_1, y_1)$ on the parabola $y^2=4ax$ is $x_1+a$. Given parabola is $y^2=kx$,so $4a=k \Rightarrow a=k/4$. The focal distance is $x_1+a = 2 + k/4 = 3$. $k/4 = 1 \Rightarrow k=4$. The parabola is $y^2=4x$. Since $P(2, y_1)$ lies on $y^2=4x$,$y_1^2 = 4(2) = 8 \Rightarrow y_1 = \pm 2\sqrt{2}$. So $P$ is $(2, 2\sqrt{2})$ or $(2, -2\sqrt{2})$. The equation of the tangent to $y^2=4x$ at $(x_1, y_1)$ is $yy_1 = 2(x+x_1)$. For $P(2, 2\sqrt{2})$: $y(2\sqrt{2}) = 2(x+2)$ $\Rightarrow \sqrt{2}y = x+2$ $\Rightarrow x - \sqrt{2}y + 2 = 0$. For $P(2, -2\sqrt{2})$: $y(-2\sqrt{2}) = 2(x+2)$ $\Rightarrow -\sqrt{2}y = x+2$ $\Rightarrow x + \sqrt{2}y + 2 = 0$. Combining these,the equation is $x \pm \sqrt{2}y + 2 = 0$.

If the focal distance of a point $P(2, y_1)$ on the parabola $y^2=kx$ is $3$,then the equation of the tangent drawn at $P$ to the given parabola is

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