If $P ( h , k )$ be a point on the parabola $x=4 y^2$, which is nearest to the point $Q (0,33)$, then the distance of $P$ from the directrix of the parabola $y^2=4(x+y)$ is equal to :
- 4
- 2
- 8
- 6
The Correct Option is D
Approach Solution - 1
We are given the equation of the parabola \(x = 4y^2\), and we need to find the distance from the point \(P(h, k)\) on the parabola to the directrix of another parabola \(y^2 = 4(x + y)\). The point \(P\) is the closest point to \(Q(0, 33)\).
Step 1: Equation of the Normal
The equation of the normal to the parabola \(x = 4y^2\) is given by:
\[ y = -tx + 2at + at^3, \]
where \(t\) is the parameter, and \(a = \frac{1}{16}\). Substituting \(a = \frac{1}{16}\), the equation of the normal becomes:
\[ y = -tx + \frac{t^2}{16} + \frac{t^3}{16}. \]
Step 2: It Passes Through \(Q(0, 33)\)
Substitute \(x = 0\) and \(y = 33\) into the normal equation:
\[ 33 = -t(0) + \frac{t^2}{16} + \frac{t^3}{16}. \]
This simplifies to:
\[ 33 = \frac{t^2}{16} + \frac{t^3}{16}. \]
Multiply through by 16:
\[ 528 = t^2 + t^3. \]
Rearranging gives:
\[ t^3 + t^2 - 528 = 0. \]
Solving this cubic equation, we find \(t = 8\).
Step 3: Parametric Coordinates of \(P\)
Now substitute \(t = 8\) into the parametric equations for \(P\) (point on the parabola):
\[ P(8, 2at) = \left(\frac{1}{16} \times 64, 2 \times \frac{1}{16} \times 8\right) = (4, 1). \]
Step 4: Parabola Equation
The equation of the given parabola is:
\[ y^2 = 4(x + y). \]
Rearranging gives:
\[ y^2 - 4y = 4x. \]
Completing the square:
\[ (y - 2)^2 = 4(x + 1). \]
Step 5: Equation of the Directrix
The equation of the directrix is:
\[ x + 1 = -1, \]
which simplifies to:
\[ x = -2. \]
Step 6: Distance from \(P\) to the Directrix
The distance of the point \(P(4, 1)\) from the directrix \(x = -2\) is given by the horizontal distance:
\[ \text{Distance} = |4 - (-2)| = 6. \]
Thus, the distance from \(P\) to the directrix is 6.
Approach Solution -2
It passes through
Parabola :
Equation of directix :-
Distance of point
Ans. : (4)
Top Questions on Parabola
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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.