Question

The headlight of a motor is a parabolic reflector of a diameter 12 cm and depth 4 cm. find the position of bulb on the axis of the reflector for effective functioning of the headlight.

Answer 1

$\text{Let PQ be the diameter of the parabolic reflector}$

$\text{Let F be the focus of the parabolic reflector}$

$\textbf{Given:}$

$PQ=12\,cm\;\text{and depth}=4\,cm$

$\textbf{To find:}$

$\text{The position of bulb on the axis of the reflector for}$

$\text{effective functioning of the headlight}$

$\textbf{Solution:}$

$\text{Let us consider the parabolic reflector as}$

open rightward

$\text{Then, the equation of the parabola is}\;y^2=4a\,x$

$\text{From the figure, it is clear that the point P is (4,6) }$

$\text{Since the point P lies on the parabola, we have}$

$6^2=4a(4)$

$36=16a$

$a=\dfrac{36}{16}$

$\implies\,a=\dfrac{9}{4}$

$\implies\,a=2.25\,cm$

$\text{That is,}\,\;VF=2.25\,cm$

$\textbf{Answer:}$

$\textbf{By the property of parabolic reflector,}$$\textbf{the bulb should be place at a distance of 2.25 cm from the vertex}$

Find more:

If the parabola y^2 = 4ax passes through the point (3, 2), then the length of its latus rectum is

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If the chord joining the points t1 and t2 to the parabola y2 = 4ax is normal to the parabola at t-th prove that t1(t1+ t2) =-2.

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Answer 2

Answer:

Let PQ be the diameter of the parabolic reflector

\text{Let F be the focus of the parabolic reflector}Let F be the focus of the parabolic reflector

\textbf{Given:}Given:

PQ=12\,cm\;\text{and depth}=4\,cmPQ=12cmand depth=4cm

\textbf{To find:}To find:

\text{The position of bulb on the axis of the reflector for}The position of bulb on the axis of the reflector for

\text{effective functioning of the headlight}effective functioning of the headlight

\textbf{Solution:}Solution:

\text{Let us consider the parabolic reflector as}Let us consider the parabolic reflector as

open rightward

\text{Then, the equation of the parabola is}\;y^2=4a\,xThen, the equation of the parabola isy

2

=4ax

\text{From the figure, it is clear that the point P is $(4,6)$}From the figure, it is clear that the point P is (4,6)

\text{Since the point P lies on the parabola, we have}Since the point P lies on the parabola, we have

6^2=4a(4)6

2

=4a(4)

36=16a36=16a

a=\dfrac{36}{16}a=

16

36

\implies\,a=\dfrac{9}{4}⟹a=

4

9

\implies\,a=2.25\,cm⟹a=2.25cm

\text{That is,}\,\;VF=2.25\,cmThat is,VF=2.25cm

\textbf{Answer:}Answer:

\textbf{By the property of parabolic reflector,}By the property of parabolic reflector, \textbf{the bulb should be place at a distance of 2.25 cm from the vertex}the bulb should be place at a distance of 2.25 cm from the vertex

Find more:

If the parabola y^2 = 4ax passes through the point (3, 2), then the length of its latus rectum is

https://brainly.in/question/8040659

If the chord joining the points t1 and t2 to the parabola y2 = 4ax is normal to the parabola at t-th prove that t1(t1+ t2) =-2.