If the shortest distance of the parabola \(y^{2}=4x\) from the centre of the circle \(x² + y² - 4x - 16y + 64 = 0\) is d, then d2 is equal to:
If the shortest distance of the parabola \(y^{2}=4x\) from the centre of the circle \(x² + y² - 4x - 16y + 64 = 0\) is d, then d2 is equal to:
16
24
36
20
The Correct Option is D
Solution and Explanation
Step 1. Rewrite the Equation of the Circle in Standard Form
Given the equation:
\(x^2 + y^2 - 4x - 16y + 64 = 0\)
Completing the square for the terms involving \(x\) and \(y\):
\((x^2 - 4x) + (y^2 - 16y) = -64\)
\((x - 2)^2 - 4 + (y - 8)^2 - 64 = -64\)
Rearranging terms:
\((x - 2)^2 + (y - 8)^2 = 4\)
Thus, the center of the circle is \((2, 8)\) and the radius is 2.
Step 2. Find the Normal to the Parabola
Consider the parabola \(y^2 = 4x\). Let the slope of the normal be \(m\). The equation of the normal to the parabola is given by:
\(y = mx - 2m - m^3\)
Substitute the point \((2, 8)\) into the equation to find \(m\):
\(8 = m \cdot 2 - 2m - m^3\)
Simplifying:
\(m^3 + 2m - 8 = 0\)
Step 3. Calculate the Distance
The shortest distance is between the center \((2, 8)\) of the circle and the point on the parabola where the normal passes. Using the distance formula, we find:
\(d^2 = (x − 2)^2 + (y − 8)^2 = 20\)
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.
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