Let a line perpendicular to the line 2x – y = 10 touch the parabola at the point P. The distance of the point P from the centre of the circle is __________. [2024]

Question

Let a line perpendicular to the line 2x – y = 10 touch the parabola $y^{2} = 4 (x - 9)$ at the point P. The distance of the point P from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________. [2024]

Smart Achievers · Accepted Answer

(10)

Let L : 2x – y = 10 and $C : y^{2} = 4 (x - 9)$

Equation of line perpendicular to L is given by

2y + x = k

Now, let us find the point of intersection of 2y + x = k and $y^{2} = 4 (x - 9)$

$i . e ., {(\frac{k - x}{2})}^{2} = 4 (x - 9)$

$\Rightarrow x^{2} - 2 (k + 8) x + (k^{2} + 144) = 0$

As parabola touches the line so this quadratic equation must have at most one real root

$\Rightarrow D = 0$

$\Rightarrow 4 {(k + 8)}^{2} - 4 (k^{2} + 144) = 0$

$\Rightarrow 16 k + 64 - 144 = 0$

$\Rightarrow k = 5$

So, equation becomes $x^{2} - 26 x + 169 = 0$

$\Rightarrow {(x - 13)}^{2} = 0$

$\Rightarrow x = 13$

$\Rightarrow y = \pm 4$

Now, parabola and line 2y + x = 5 meets at P(13, –4)

Now, centre of given circle is (7, 4)

$∴$ Required distance $= \sqrt{{(13 - 7)}^{2} + {(- 4 - 4)}^{2}} = 10$ .

Solution

Q.
Let a line perpendicular to the line 2x – y = 10 touch the parabola $y^{2} = 4 (x - 9)$ at the point P. The distance of the point P from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________. [2024]

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Solution

Q. Let a line perpendicular to the line 2x – y = 10 touch the parabola y2=4(x–9) at the point P. The distance of the point P from the centre of the circle x2+y2–14x–8y+56=0 is __________. [2024]

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Q.
Let a line perpendicular to the line 2x – y = 10 touch the parabola $y^{2} = 4 (x - 9)$ at the point P. The distance of the point P from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________. [2024]