If the focal chord drawn through the point P( | vedclass

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(C) The parabola is $y^2=5x$,so $4a=5 \Rightarrow a=\frac{5}{4}$. The focus $S$ is $\left(\frac{5}{4}, 0\right)$.The focal chord passes through $P(5,5

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$\left(\frac{-5}{16}, 0\right)$

Vedclass · Answer

(C) The parabola is $y^2=5x$,so $4a=5 \Rightarrow a=\frac{5}{4}$. The focus $S$ is $\left(\frac{5}{4}, 0\right)$.The focal chord passes through $P(5,5)$ and $S\left(\frac{5}{4}, 0\right)$.The slope of the chord $PS$ is $m = \frac{5-0}{5-\frac{5}{4}} = \frac{5}{15/4} = \frac{20}{15} = \frac{4}{3}$.The equation of the focal chord is $y-0 = \frac{4}{3}\left(x-\frac{5}{4}\right)$ $\Rightarrow 3y = 4x-5$ $\Rightarrow 4x-3y=5$.To find $Q$,substitute $x = \frac{3y+5}{4}$ into $y^2=5x$:$y^2 = 5\left(\frac{3y+5}{4}\right)$ $\Rightarrow 4y^2 - 15y - 25 = 0$ $\Rightarrow (4y+5)(y-5) = 0$.Since $P$ is $(5,5)$,$Q$ must be $\left(\frac{5}{16}, -\frac{5}{4}\right)$.The tangent at $Q(x_1, y_1)$ is $yy_1 = \frac{5}{2}(x+x_1)$.Substituting $Q\left(\frac{5}{16}, -\frac{5}{4}\right)$: $y\left(-\frac{5}{4}\right) = \frac{5}{2}\left(x+\frac{5}{16}\right) \Rightarrow -\frac{1}{2}y = x+\frac{5}{16}$.The axis of the parabola is $y=0$. Setting $y=0$ in the tangent equation gives $x = -\frac{5}{16}$.Thus,the point is $\left(-\frac{5}{16}, 0\right)$.

If the focal chord drawn through the point $P(5,5)$ to the parabola $y^2=5x$ meets the parabola again at the point $Q$,then the tangent drawn to this parabola at $Q$ meets the axis of the parabola at the point

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