Inter 1st Year Maths Conic Sections Solutions Exercise 10e

Practicing the AP Board Solutions Class 11 Maths and Chapter 10 Inter 1st Year Maths Conic Sections Solutions Exercise 10e Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Conic Sections Solutions Exercise 10e

Conic Sections Exercise 10e Solutions

Conic Sections Class 11 Exercise 10e Solutions – Conic Sections 10e Exercise Solutions

I.

Question 1.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Solution:
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.
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),
10² = 4a(5)
⇒ 100 = 20a
⇒ a = 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 midpoint of the diameter.

Question 2.
An arch is in the form of a semi-ellipse. It is 8m wide and 2m high at the centre. Find the height of the arch at a point 1.5m from one end.
Solution:
Since the height and width of the arc from the centre are 2m and 8m respectively.
The length of the major axis is 8m, while the length of the semi-minor axis is 2m.
The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis.
Hence, the semi-ellipse can be diagrammatically represented as
The equation of the semi-ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), y ≥ 0
where a is the semi-major axis

Accordingly, 2a = 8
⇒ a = 4, b = 2
Therefore, the equation of the semi-ellipse is \(\frac{x^2}{16}+\frac{y^2}{4}=1\), y ≥ 0 ………..(1)
Let A be a point on the major axis
such that AB = 1.5m
Draw AC ⊥ OB
OA = (4 – 1.5)m = 2.5m
The x-coordinate of point C is 2.5.
On substituting the value of x with 2.5 in equation (1), we obtain

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56m.

II.

Question 1.
An arch is in the form of a parabola with its axis vertical. The arch is 10m high and 5m wide at the base. How wide is it, 2m from the vertex of the parabola?
Solution:
The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the positive y-axis.
The equation of the parabola is of the form x² = 4ay (as it is opening upwards)
It can be clearly seen that the parabola passes through the point (\(\frac {5}{2}\), 10)
Therefore, the arch is in the form of a parabola whose equation is x = -y
∴ \(\left(\frac{5}{2}\right)^2\) = 4a(10)
⇒ a = \(\frac{25}{4 \times 4 \times 10}=\frac{5}{32}\)

Therefore, the arch is in the form of a parabola whose equation is x² = \(\frac {5}{8}\)y
When y = 2m, x² = \(\frac {5}{8}\) × 2
⇒ x² = \(\frac {5}{4}\)
⇒ x = \(\sqrt{\frac{5}{4}}\)m
∴ AB = 2 × \(\sqrt{\frac{5}{4}}\)m
= 2 × 1.118 m (approx)
= 2.23 m (approx)
Hence, when the arch is 2m from the vertex of the parabola, its width is approximately 2.23 m.

Question 2.
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway, which is horizontal and 100 m long, is supported by vertical wires attached to the cable, the longest wire being 30m and the shortest being 6m. Find the length of a supporting wire attached to the roadway 18m from the middle.
Solution:
The vertex is at the lowest point of the cable.
The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis.
This can be diagrammatically represented as in the figure.

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.
DF is the supporting wire attached to the roadway, 18 m from the middle.
Here, AB = 30 m, OC = 6m and BC = \(\frac {100}{2}\) = 50 m.
The equation of the parabola is of the form x² = 4ay (as it is opening upwards)
The coordinates of point A are (50, 30 – 6) = (50, 24)
Since A(50, 24) is a point on the parabola,
(50)² = 4a(24)
⇒ a = \(\frac{50 \times 50}{4 \times 24}=\frac{625}{24}\)
∴ Equation of the parabola, x² = 4 × \(\frac {625}{24}\) × y
⇒ 6x² = 625y
The x-coordinate of point D is 18.
Hence, at x = 18, 6(18)² = 625y
⇒ y = \(\frac{6 \times 18 \times 18}{625}\)
⇒ y = 3.11 (approx.)
∴ DE = 3.11 m
DF = DE + EF
= 3.11 m + 6 m
= 9.11 m
Thus, the length of the supporting wire attached to the roadway, 18m from the middle, is approximately 9.11m.

Question 3.
A rod of length 12 cm moves with its end always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.
Solution:
Let AB be the rod making an angle θ with OX, and P(x, y) be the point on it such that AP = 3 cm.
Then, PB = AB – AP
= (12 – 3) cm
= 9 cm [AB = 12 cm]
From P, draw PQ ⊥ OY, and PR ⊥ OX

In ∆PBQ, cos θ = \(\frac{P Q}{P B}=\frac{x}{9}\)
In ∆PRA, sin θ = \(\frac{P R}{P A}=\frac{x}{3}\)
Since, sin²θ + cos²θ = 1
⇒ \(\left(\frac{y}{3}\right)^2+\left(\frac{x}{9}\right)^2=1\)
⇒ \(\frac{x^2}{81}+\frac{y^2}{9}=1\)
Thus, the equation of the locus of point P on the rob is \(\frac{x^2}{81}+\frac{y^2}{9}=1\).

III.

Question 1.
Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of its latus rectum.
Solution:
The given parabola is x² = 12y.
On comparing this equation with x² = 4ay,
we obtain 4a = 12
⇒ a = 3
∴ The coordinates of foci are S(0, a) = S(0, 3).

Let AB be the latus rectum of the given parabola.
The given parabola can be roughly drawn as in the figure.
At y = 3, x² = 12(3)
⇒ x² = 36
⇒ x = ±6
The coordinates of A are (-6, 3), while the coordinates of B are (6, 3).
Therefore, the vertices of ∆OAB are O(0, 0), A(-6, 3) and B(6, 3).
Area of ∆OAB = \(\frac {1}{2}\)|6 × 3 + 6 × 3|
= 18 sq units.

Question 2.
A man running a racecourse notes that the sum of the distances from the two flag posts to him is always 10m and the distance between the flag posts is 8 m. Find the equation of the path traced by the man.
Solution:
A man is running such that the sum of his distances from two fixed flag posts is always 10 m = (foci)
The distance between the two flag posts is 8m = foci
F₁ = (-4, 0) and F₂ = (4, 0)
This places the foci 8 meters apart along the x-axis (since 4 + 4 = 8 m)
Let man be at point (x, y)
d₁ = \(\sqrt{(x+4)^2+y^2}\)
d₂ = \(\sqrt{(x-4)^2+y^2}\)
∴ d₁ + d₂ = 10
For an ellipse with foci at (-c, 0) and (c, 0)
Sum of distances from any point to the foci = 2a
⇒ 2a = 10
⇒ a = 5
Distance between foci = 2c = 8
⇒ c = 4
b² = a² – c²
= 5² – 4²
= 25 – 16
= 9
Using the standard form of an ellipse centred at the origin \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
⇒ \(\frac{x^2}{25}+\frac{y^2}{9}=1\)
∴ Major axis = 10 m
∴ Minor aixs = 6 m
∴ Foci 8m apart along the x-axis.

Question 3.
An equilateral triangle is inscribed in the parabola y² = 4ax, where one vertex is the vertex of the parabola. Find the length of the side of the triangle.
Solution:
Let’s assume:
The vertex of the parabola is O(0, 0), and this is one vertex of the triangle.
The triangle is equilateral and inscribed in the parabola y² = 4ax.
A = (0, 0), B = (x₁, y₁), C = (x₂, y₂)
\(y_1^2=4 a x_1\) and \(y_2^2=4 a x_2\)
If B and C are symmetric across the X-axis and A is the origin,
The triangle ABC becomes equilateral if the angle between AB and AC is 60°.
A = (0, 0), B = (x, y), C = (x, -y)
AB² = x² + y²
AC² = x² + y²
BC² = (x – x)² + (y + y)²
= 0 + (2y)²
= 4y²
For a triangle to be equilateral,
AB² = BC²
⇒ x² + y² = 4y²
⇒ x² = 3y² ……(1)
Also, since B lies on the parabola y² = 4ax
⇒ x = \(\frac{y^2}{4 a}\) ………(2)
From (1), x² = 3y²
⇒ \(\left(\frac{y^2}{4 a}\right)^2\) = 3y² (by (2))
⇒ \(\frac{y^4}{16 a^2}\) = 3y2
⇒ y⁴ = 48a²y²
⇒ y² = 48a²
⇒ y = \(\sqrt{48 a^2}\)
⇒ y = 4√3a
Now find x from (2)
x = \(\frac{y^2}{4 a}=\frac{48 a^2}{4 a}\) = 12a
AB² = x² + y²
= (12a)² + (4√3a)²
= 144a² + 48a²
= 192a²
⇒ S = \(\sqrt{192 \mathrm{a}^2}\) = 8√3a
∴ Side length of the triangle = 8√3a.