If the focal distance of a point on the parab | vedclass

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(C) The given parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = 1/4$.Let the point on the parabola be $P(x_1, y_1)$. The

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$4x + y = 18$

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(C) The given parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = 1/4$.Let the point on the parabola be $P(x_1, y_1)$. The focal distance of a point $P(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $x_1 + a$.Given $x_1 + 1/4 = 17/4$,we get $x_1 = 16/4 = 4$.Since $y^2 = x$,$y^2 = 4$,so $y = \pm 2$. Since the point is in the first quadrant,$y_1 = 2$.Thus,the point is $P(4, 2)$.The slope of the tangent at $(x_1, y_1)$ is $m_t = \frac{1}{2y_1} = \frac{1}{2(2)} = 1/4$.The slope of the normal $m_n$ is $-1/m_t = -4$.The equation of the normal at $(4, 2)$ is $y - 2 = -4(x - 4)$.$y - 2 = -4x + 16$.$4x + y = 18$.

If the focal distance of a point on the parabola $y^2 = x$ is $17/4$,and it lies in the first quadrant,find the equation of the normal at this point.

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