RS Aggarwal Solutions for Class 11 Chapter 25 Applications of Conic Sections

Question 1. The focus of a parabolic mirror is at a distance of 6 cm from its vertex. If the mirror is 20 cm deep, find its diameter.
Answer: The focus sits 6 cm away from the vertex of a parabolic mirror. The mirror has a depth of 20 cm. We need to calculate the mirror's diameter.

Let O be the vertex and A be the focus, so OA = a = 6 cm. OD represents the mirror's depth = 20 cm, and BC represents the mirror's diameter.

The parabola's equation is: \( y^2 = 4ax \)

\( \Rightarrow y^2 = 24x \)

Since the mirror extends 20 cm deep, points B, C, and D all have an x-coordinate of 20.

Both points B and C lie on the parabola, so they satisfy the equation:

\( y^2 = 24 \times 20 = 480 \)

\( \Rightarrow y = \pm 21.9 \)

Point B has coordinates (20, 21.9) and point C has coordinates (20, -21.9).

Therefore, the mirror's diameter = (21.9 + 21.9) cm = 43.8 cm
In simple words: A parabolic mirror follows a specific curve equation. When you plug in how deep the mirror is, you find how wide it spreads at that depth. Adding the two halves of the width gives you the total diameter.

Exam Tip: Always set up the parabola equation first using the focus distance, then substitute the depth value to find the width at that point. Verify both y-values are symmetric.

Question 2. A parabolic reflector is 5 cm deep and its diameter is 20 cm. How far is its focus from the vertex?
Answer: A parabolic reflector measures 5 cm in depth with a diameter of 20 cm. We must find how far the focus is located from the vertex.

The reflector's depth is OD = 5 cm, and the mirror's diameter is BC = 20 cm.

Suppose the parabola's equation is \( y^2 = 4ax \), where a represents the distance from the focus to the vertex.

Points B and C both have an x-coordinate of 5. Point D is the midpoint of BC on the x-axis, so BD = CD = 10 cm.

Thus, point B has coordinates (5, 10).

Substituting into the equation:

\( y^2 = 4ax \)

\( \Rightarrow 100 = 4a \times 5 \)

\( \Rightarrow 20a = 100 \)

\( \Rightarrow a = 5 \)

Therefore, the focus is 5 cm away from the vertex.
In simple words: Using the width and depth of the reflector, we can work backwards through the parabola equation to find where the focus must be positioned.

Exam Tip: Recognize that the reflector's diameter and depth give you a point on the parabola; use that point to solve for the parameter a, which is the focal distance.

Question 3. The towers of bridge, hung in the form of a parabola, have their tops 30 m above the roadway, and are 200 m apart. If the cable is 5 m above the roadway at the center of the bridge, find the length of the vertical supporting cable, 30 m from the center.
Answer: The bridge's towers sit 30 m above the roadway and are spaced 200 m apart. At the bridge's center, the cable hangs 5 m above the roadway. We must calculate the vertical cable length 30 m away from the center.

Points A and B mark the tower tops. AE and BF show the tower heights. H is the bridge's center, and HI measures 5 m up from the roadway.

Suppose the parabola's equation is: \( x^2 = 4a(y - b) \)

Here, b = 5, so: \( x^2 = 4a(y - 5) \)

Since AB = 200 m and BF = 30 m, point B has coordinates (100, 30).

Point B lies on the parabola, so:

\( x^2 = 4a(y - 5) \)

\( \Rightarrow 10000 = 4a(30 - 5) \)

\( \Rightarrow 10000 = 4a \times 25 \)

\( \Rightarrow a = 100 \)

To find the cable length 30 m from the center, the x-coordinate is 30:

\( 30 \times 30 = 4a(y - 5) \)

\( \Rightarrow 900 = 400(y - 5) \)

\( \Rightarrow y - 5 = \frac{9}{4} \)

\( \Rightarrow y = \frac{9}{4} + 5 = \frac{29}{4} = 7.25 \) m

Therefore, the vertical supporting cable length 30 m from the center is 7.25 m.
In simple words: Bridge cables form a parabolic curve. By knowing the tower heights and center sag, you can find the cable height at any distance from the middle.

Exam Tip: Identify the vertex height carefully (here, 5 m above ground) and adjust the parabola equation accordingly. Always verify your coordinate substitutions match the given distances.

Question 4. A rod of length 15 cm moves with its ends always touching the coordinate axes. Find the equation of the locus of a point P on the rod, which is at a distance of 3 cm from the end in contact with the x-axis.
Answer: A rod measuring 15 cm in length slides such that its endpoints remain on the coordinate axes. A point P on the rod sits 3 cm away from the endpoint touching the x-axis. We must find the locus equation for point P.

Rod AB makes an angle θ with the x-axis. The distance AP = 3 cm.

Therefore, PB = AB - AP = 15 - 3 = 12 cm

PQ is the perpendicular drawn to the x-axis, and RP is the perpendicular drawn to the y-axis.

Let point P have coordinates (x, y).

In triangle ΔBPQ:

\( \cos\theta = \frac{x}{PB} = \frac{x}{9} \)

In triangle ΔPAR:

\( \sin\theta = \frac{y}{AP} = \frac{y}{3} \)

Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):

\( \left(\frac{y}{3}\right)^2 + \left(\frac{x}{9}\right)^2 = 1 \)

\( \Rightarrow \frac{x^2}{81} + \frac{y^2}{9} = 1 \)

This is the locus of point P.
In simple words: As the rod slides with its ends on both axes, a fixed point on the rod traces out an ellipse. The equation shows how x and y coordinates relate for every position the rod takes.

Exam Tip: Use trigonometric ratios to express coordinates in terms of the angle θ, then apply the fundamental identity to eliminate θ and obtain the locus equation.

Question 5. A beam is supported at its ends by supports which are 12 m apart. Since the load is concentrated at its center, there is a deflection of 3 cm at the center, and the deflected beam is in the shape of a parabola. How far from the center is the deflection 1 cm?
Answer: A beam rests on supports positioned 12 m apart. Because the load concentrates at the center, the beam sags 3 cm at its midpoint, forming a parabolic shape. We must find the horizontal distance from center where the sag measures 1 cm.

Points E and F mark the beam's ends, separated by 12 m. Point I marks the center with a 3 cm sag.

The distance IF = 12/2 = 6 m = 600 cm, and the deflection IJ = FH = 3 cm.

Point F has coordinates (600, 3).

The parabola's equation is: \( x^2 = 4ay \)

Since F lies on the parabola:

\( x^2 = 4ay \)

\( \Rightarrow 3600 = 4a \times 3 \)

\( \Rightarrow a = 300 \)

Point KL denotes the 1 cm deflection. At point L, the y-coordinate is (3 - 1) = 2.

Using the equation:

\( x^2 = 4ay = 4 \times 300 \times 2 = 2400 \)

\( \Rightarrow x = 49 \) cm

Therefore, the distance from center where the deflection equals 1 cm is 49 cm.
In simple words: A sagging beam curves parabolically. Once you know the maximum sag at the center, you can calculate how much the beam sags at any other distance along its length.

Exam Tip: Remember that deflection is measured downward from the original beam position, so smaller deflection means a y-value closer to zero. Always adjust y-coordinates accordingly.