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 \[ x^2 + y^2 - 14x - 8y + 56 = 0 \] is _____.
Correct Answer: 10
Solution and Explanation
Given:
\( y^2 = 4(x-9) \),
which represents a parabola with vertex at \( (9,0) \) and axis along the \( x \)-axis.
Step 1: Finding the Slope of the Perpendicular Line The given line is:
\( 2x - y = 10. \)
Rearranging:
\( y = 2x - 10, \)
with a slope of \( 2 \). A line perpendicular to this has a slope:
\( m = -\frac{1}{2}. \)
Step 2: Equation of the Tangent The equation of the tangent to the parabola \( y^2 = 4(x-9) \) at a point \( (x_1, y_1) \) is given by:
\( yy_1 = 2(x + x_1 - 9). \)
Substituting the slope \( m = -\frac{1}{2} \) into the equation of the tangent:
\( y = -\frac{1}{2}x + c. \)
Equating with the general form and solving for the point of contact \( P \), we find:
\( P(13, -4). \)
Step 3: Centre of the Circle Given the equation of the circle:
\( x^2 + y^2 - 14x - 8y + 56 = 0. \)
Completing the square:
\( (x-7)^2 + (y-4)^2 = 9. \)
The centre of the circle is:
\( C(7, 4). \)
Step 4: Calculating the Distance \( CP \) The distance between point \( P(13, -4) \) and the centre \( C(7, 4) \) is given by:
\( CP = \sqrt{(13-7)^2 + (-4-4)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10. \)
Therefore, the distance \( CP \) is \( 10 \).
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