If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Solution and Explanation
The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis.
This can be diagrammatically represented as
The equation of the parabola is of the form y2 = 4ax (as it is opening to the right). Since the parabola passes through point A (5, 10),
\(10^2 = 4a(5)\)
\(⇒ 100 = 20a\)
\(⇒ a = \frac{100}{20} = 5\)
Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter.
Hence, the focus of the reflector is at the mid-point of the diameter.
Top Questions on Parabola
- Let \( L_1, L_2 \) be the lines passing through the point \( P(0, 1) \) and touching the parabola \[ 9x^2 + 12x + 18y - 14 = 0. \] Let \( Q \) and \( R \) be the points on the lines \( L_1 \) and \( L_2 \) such that the \( \Delta PQR \) is an isosceles triangle with base \( QR \). If the slopes of the lines \( QR \) are \( m_1 \) and \( m_2 \), then \( 16\left(m_1^2 + m_2^2\right) \) is equal to ______.
- Let \( C \) be the circle of minimum area touching the parabola \( y = 6 - x^2 \) and the lines \( y = \sqrt{3} |x| \). Then, which one of the following points lies on the circle \( C \)?
- Let the length of the focal chord \( PQ \) of the parabola \( y^2 = 12x \) be 15 units. If the distance of \( PQ \) from the origin is \( p \), then \( 10p^2 \) is equal to _____
- Let \( PQ \) be a chord of the parabola \( y^2 = 12x \) and the midpoint of \( PQ \) be at \( (4, 1) \). Then, which of the following points lies on the line passing through the points \( P \) and \( Q \)?
- The area (in square units) of the region described by: \[ \{(x, y) : y^2 \leq 2x, \, \text{and} \, y \geq 4x - 1\} \] is:
Questions Asked in CBSE Class XI exam
- Draw formulas for the first five members of each homologous series beginning with the following compounds.\((a) H–COOH (b) CH_3COCH_3 (c) H–CH=CH_2\)
- CBSE Class XI
- Organic Chemistry- Some Basic Principles and Techniques
- Distinguish between the following pairs of sentences.
(i) He was visibly moved.
(ii) He was visually impaired.- CBSE Class XI
- The adventure
- How, according to you, can peace and liberty be maintained in a state?
- CBSE Class XI
- The tale of melon City
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
- CBSE Class XI
- Pressure
- Describe the change in hybridisation (if any) of the Al atom in the following reaction.
\(AlCl_3+Cl^- \rightarrow AlCl_4^-\)- CBSE Class XI
- Hybridisation
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.