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

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Aicharya · Accepted Answer

$\because y^2=5 x \Rightarrow$ focus $\equiv\left(\frac{5}{4}, 0\right)$ Equation of focal chord

$ (y-5)=\left(\frac{5-0}{5-\frac{5}{4}}\right)(x-5) \Rightarrow 4 x-3 y=5 $ Solving $\mathrm{y}^2=5 x \& 4 x-3 y=5$, we get: Coordinate of $Q \equiv\left(\frac{5}{16},-\frac{5}{4}\right)$ Slope at point $Q \Rightarrow m=\frac{d y}{d x}=\frac{S}{2 y}=-2$ Now, the equation of tangent at $Q$ is $y+\frac{5}{4}=-2\left(x-\frac{5}{16}\right) \Rightarrow y+\frac{5}{4}=-2 x+\frac{10}{8}$ ... (i) Now, since the axis of parabola is $y=0$ So, from equation (i), $x=-\frac{5}{16}$ $\therefore$ The required point is $\left(\frac{5}{16}, 0\right)$

If the focal chord drawn through the point $\mathrm{P}(5,5)$ to the parabola $\mathrm{y}^2=5 \mathrm{x}$…

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