If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

ncert cbse 11th mathematics chapter 11 conic section 

miscellaneous

 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus

 

radius at the opening is 20/2 = 10

Choose the parabola as y^2 = 4ax

(5,10) lies on the parabola so that

10^2 = 4a(5)

or a = 100/20

a = 5

Therefore focus and centre are at  of 5 cm apart

In other words the focus is the midpoint of the given diameter.

 

2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high
and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Choosing the parabola as (x^2)  = -4ay open downwards

because base is 5m wide at the base, (5/2) =  2.5m to each side  at 10 m depth

we can assume that (2.5,-10) lies on the parabola  (x^2)  = -4ay

so that (2.5) ^2 = -4 * a *(-10)

6.25 = 40 a

a =6.25 /40

resubstitute in (x^2)  = -4(6.25 /40)y

to get  (x^2)  = ( - 0.625 ) y

(x^2)  = ( - 0.625 )y

at 2 m distant, let (x,-2)  lie on (x^2)  = -40y

 (x^2)  = ( - 0.625 )(-2)

(x^2)  = 1.25

 (x^2)  = 125 / 100

 (x^2)  = 5 / 4

x =(sqrt(5)) / 2

Required width  = 2 * [(sqrt(5)) / 2] = sqrt(5)

Required width =2.23 m approximately

 

 

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ncert cbse 11th mathematics chapter 11 conic section 

miscellaneous

 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus

solution

2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high
and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

solution 

 

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

 

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 13. Find the equation of the right bisector of the line segment joining the points

 (3, 4) and (–1, 2).

solution

 

10. The line through the points (h, 3) and (4, 1) intersects the line 

7 x − 9 y − 19 = 0 at right angle. Find the value of h. 

solution

 

 

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having
x intercept 3.

solution  

 

exercise 10.2

19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find
equation of the line.

solution

17.The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear
relationship between selling price and demand, how many litres could he sell
weekly at Rs 17/litre?

solution

13. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.  

solution 

 

12.Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

 solution

 

11.A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.   

 solution 

10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6). 

solution

 

9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the
median through the vertex R.

solution

 

 

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