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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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Verified Answer
The correct answer is:
$\left(\frac{-5}{16}, 0\right)$
$\because y^2=5 x \Rightarrow$ 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 $\left(\frac{5}{16}, 0\right)$
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 $\left(\frac{5}{16}, 0\right)$
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