# CLASS 11 MATHS CHAPTER-11 CONIC SECTIONS

### In each of the following Exercises 1 to 5, find the equation of the circle with:

Question1. Find the equation of the circle with centre (0, 2) and radius 2

Solution :
The equation of a circle with centre (h, k) and radius r is given as

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)2 + (y – 2)2 = 22

x2 + y2 + 4 – 4 y = 4

x2 + y2 – 4y = 0

Question2. Find the equation of the circle with centre (–2, 3) and radius 4

Solution :
The equation of a circle with centre (h, k) and radius r is given as

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is

(x + 2)2 + (y – 3)2 = (4)2

x2 + 4x + 4 + y2 – 6y + 9 = 16

x2 + y2 + 4x – 6y – 3 = 0

Solution :

The equation of a circle with centre (hk) and radius is given as

(x­ – h)2 + (y ­– k)2 = r2

It is given that centre (hk) = and radius (r) =.

Therefore, the equation of the circle is

Question4. Find the equation of the circle with centre (1,1) and radius 2

Solution :
The equation of a circle with centre (h, k) and radius r is given as

(x−h)2 + (y−k)= r2

It is given that centre (h, k) = (1, 1) and radius (r) = √

Therefore, the equation of the circle is

(x−1)2+(y−1)2 = (√2)2x2 − 2x + 1 + y2 − 2y + 1

= 2x2 + y2 − 2x− 2y=0

Question5. Centre (–a, –b) and radius

Solution :

The equation of a circle with centre (hk) and radius is given as

(x­ – h)2 + (y ­– k)2 = r2

It is given that centre (hk) = (–a, –b) and radius (r) =.

Therefore, the equation of the circle is

In each of the following Exercises 6 to 9, find the centre and radius of the circles.

Question6. Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36

Solution :

The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.

(x + 5)2 + (y – 3)2 = 36

⇒ {x – (–5)}2 + (y – 3)2 = 62, which is of the form (x – h)2 + (y – k)2 = r2, where h = –5, k = 3, and r = 6.

Thus, the centre of the given circle is (–5, 3), while its radius is 6.

Question7. Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0

Solution :
The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0.

x2 + y– 4x – 8y – 45 = 0

⇒ (x2 – 4x) + (y2 – 8y) = 45

⇒ {x2 – 2(x)(2) + 22} + {y2 – 2(y)(4)+ 42} – 4 –16 = 45

⇒ (x – 2)2 + (y –4)2 = 65

⇒ (x – 2)2 + (y –4)2

, which is of the form (x – h)2 + (y – k)2 = r2, where h = 2, k = 4, and .

Thus, the centre of the given circle is (2, 4), while its radius .

Question8. Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0

Solution :
The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0.

x2 + y2 – 8x + 10y – 12 = 0

⇒ (x2 – 8x) + (y2 + 10y) = 12

⇒ {x– 2(x)(4) + 42} + {y2 + 2(y)(5) + 52}– 16 – 25 = 12

⇒ (x – 4)2 + (y + 5)2 = 53

, which is of the form (x – h)2 + (y – k)2 = r2, where h = 4, k = –5, and .

Thus, the centre of the given circle is (4, –5), while its radius is.

Question 9.Find the centre and radius of the circle 2x2 + 2y2 – x = 0

Solution :
The equation of the given circle is 2x2 + 2y2 – x = 0.

, which is of the form (x – h)2 + (y – k)2 = r2, where h = 1/4, k = 0, and r = 1/4.

Thus, the centre of the given circle is, while its radius is 1/4.

Question10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Solution :
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through points (4, 1) and (6, 5),

(4 – h)2 + (1 – k)2 = r2 … (1)

(6 – h)2 + (5 – k)2 = r2 … (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k = 16 … (3)

From equations (1) and (2), we obtain

(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2

⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h+ 25 – 10k + k2

⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

⇒ 4h + 8k = 44

⇒ h + 2k = 11 … (4)

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)2 + (1 – 4)2 = r2

⇒ (1)2 + (– 3)2 = r2

⇒ 1 + 9 = r2

⇒ r2 = 10

Thus, the equation of the required circle is

(x – 3)2 + (y – 4)2 =

x2 – 6x + 9 + y2 – 8y + 16 = 10

x2 + y2 – 6x – 8y + 15 = 0

Question11. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

Solution :

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through points (2, 3) and (–1, 1),

(2 – h)2 + (3 – k)2 = r2 … (1)

(–1 – h)2 + (1 – k)2 = r2 … (2)

Since the centre (h, k) of the circle lies on line – 3y – 11 = 0,

– 3k = 11 … (3)

From equations (1) and (2), we obtain

(2 – h)+ (3 – k)2 = (–1 – h)2 + (1 – k)2

⇒ 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 1 – 2k + k2

⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

⇒ 6h + 4k = 11 … (4)

On solving equations (3) and (4), we obtain .

On substituting the values of h and k in equation (1), we obtain

Thus, the equation of the required circle is

Question12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Solution :

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x – h)2 + y2 = 25.

It is given that the circle passes through point (2, 3).

When h = –2, the equation of the circle becomes

(x + 2)2 + y2 = 25

x2 + 4+ 4 + y2 = 25

x2 + y2 + 4x – 21 = 0

When h = 6, the equation of the circle becomes

(x – 6)y2 = 25

x2 – 12x +36 + y2 = 25

x2 + y2 – 12x + 11 = 0

Question13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Solution :
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.

Since the circle passes through (0, 0),

(0 – h)2 + (0 – k)2 = r2

⇒ h2 + k2 = r2

The equation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

(a – h)2 + (0 – k)2 = h2 + k2 … (1)

(0 – h)2 + (b – k)2 = h2 + k2 … (2)

From equation (1), we obtain

a– 2ah + h2 + k2 = h2 + k2

⇒ a2 – 2ah = 0

⇒ a(a – 2h) = 0

⇒ a = 0 or (a – 2h) = 0

However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h = a/2.

From equation (2), we obtain

h+ b2 – 2bk + k2 = h2 + k2

⇒ b2 – 2bk = 0

⇒ b(b – 2k) = 0

⇒ b = 0 or (b – 2k) = 0

However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k = b/2.

Thus, the equation of the required circle is

Question14. Find the equation of the circle with centre (2, 2) and passes through the point (4, 5).

Solution :

The centre of the circle is given as (hk) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

Thus, the equation of the circle is

Question15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Solution :
The equation of the given circle is x2 + y= 25.

x2 + y2 = 25

⇒ (x – 0)2 + (y – 0)2 = 52, which is of the form (x – h)2 + (y – k)2 = r2, where h = 0, k = 0, and r = 5.

∴Centre = (0, 0) and radius = 5

Distance between point (–2.5, 3.5) and centre (0, 0)

Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.

### In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

Question1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x

Solution :
The given equation is y2 = 12x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y2 = 4ax, we obtain

4a = 12 ⇒ a = 3

∴Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12

Question2. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = 6y

Solution :
The given equation is x2 = 6y.

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

On comparing this equation with x2 = 4ay, we obtain

∴Coordinates of the focus = (0, a) =

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 6

Question 3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = – 8x

Solution :
The given equation is y2 = –8x.

Here, the coefficient of x is negative. Hence, the parabola opens towards the left.

On comparing this equation with y2 = –4ax, we obtain

–4a = –8 ⇒ a = 2

∴Coordinates of the focus = (–a, 0) = (–2, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of directrix, x = a i.e., x = 2

Length of latus rectum = 4a = 8

Question 4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = – 16y

Solution :

The given equation is x2 = –16y.

Here, the coefficient of is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = – 4ay, we obtain

–4a = –16 ⇒ a = 4

∴Coordinates of the focus = (0, –a) = (0, –4)

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix, y = a i.e., y = 4

Length of latus rectum = 4a = 16

Question 5. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 10x

Solution :

The given equation is y2 = 10x.

Here, the coefficient of is positive. Hence, the parabola opens towards the right.

On comparing this equation with y= 4ax, we obtain

∴Coordinates of the focus = (a, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of directrix,

Length of latus rectum = 4a = 10

Question6. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = –9y

Solution :

The given equation is x2 = –9y.

Here, the coefficient of is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = –4ay, we obtain

∴Coordinates of the focus =

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 9

In each of the following Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

Question7. Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6

Solution :
Focus (6, 0); directrix, x = –6

Since the focus lies on the x-axis, the x-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form y2 = 4ax or

y2 = – 4ax.

It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis. Hence, the parabola is of the form y= 4ax.

Here, a = 6

Thus, the equation of the parabola is y2 = 24x.

Question8. Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3

Solution :
Focus = (0, –3); directrix y = 3

Since the focus lies on the y-axis, the y-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form x2 = 4ay or x2 = – 4ay.

It is also seen that the directrix, y = 3 is above the x-axis, while the focus

(0, –3) is below the x-axis. Hence, the parabola is of the form x2 = –4ay.

Here, a = 3

Thus, the equation of the parabola is x2 = –12y.

Question9. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0)

Solution :
Vertex (0, 0); focus (3, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y2 = 4ax.

Since the focus is (3, 0), a = 3.

Thus, the equation of the parabola is y2 = 4 × 3 × x,

i.e., y2 = 12x

Question10. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)

Solution :
Vertex (0, 0) focus (–2, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y= –4ax.

Since the focus is (–2, 0), a = 2.

Thus, the equation of the parabola is y2 = –4(2)x,

i.e., y2 = –8x

Question11. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis

Solution :

Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form y2 = 4ax or y2 = –4ax.

The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form y2 = 4ax, while point

(2, 3) must satisfy the equation y2 = 4ax.

Thus, the equation of the parabola is

Question12. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis

Solution :
Given:

Since the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form x2 = 4ay or x2 = –4ay.

The parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form x2 = 4ay, while point

(5, 2) must satisfy the equation x2 = 4ay.

Thus, the equation of the parabola is

### In each of the Exercises 1 to 6, find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

Question 1.

Solution :

The given equation is.

On comparing this equation with the standard equation of hyperbola i.e.,, we obtain a = 4 and b = 3.

We know that a2 + b2 = c2.

Therefore,

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

Length of latus rectum

Question 2.

Solution :

The given equation is.

On comparing this equation with the standard equation of hyperbola i.e.,, we obtain a = 3 and .

We know that a2 + b2 = c2.

Therefore,

The coordinates of the foci are (0, ±6).

The coordinates of the vertices are (0, ±3).

Length of latus rectum

Question 3.9 y2−4x2 = 36

Solution :

The given equation is 9y2 – 4x2 = 36.

It can be written as

9y2 – 4x2 = 36

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a = 2 and b = 3.

We know that a2 + b2 = c2.

Therefore,

The coordinates of the foci are.

The coordinates of the vertices are.

Length of latus rectum

Question 4. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x2 – 9y2 = 576

Solution :

The given equation is 16x2 – 9y2 = 576.

It can be written as

16x2 – 9y2 = 576

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a = 6 and b = 8.

We know that a2 + b2 = c2.

Therefore,

The coordinates of the foci are (±10, 0).

The coordinates of the vertices are (±6, 0).

Length of latus rectum

Question 5.5 y2 – 9x2 = 36

Solution :

The given equation is 5y2 – 9x2 = 36.

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a =  and b = 2.

We know that a2 + b2 = c2.

Therefore, the coordinates of the foci are.

The coordinates of the vertices are.

Length of latus rectum

Question 6. 49 x2 = 784

Solution :

The given equation is 49y2 – 16x2 = 784.

It can be written as
49y2 – 16x2 = 784

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a = 4 and b = 7.

We know that a2 + b2 = c2.

Therefore,

The coordinates of the foci are.

The coordinates of the vertices are (0, ±4).

Length of latus rectum

In each of the Exercises 7 to 15, find the equation of the hyperbola satisfying the given conditions.

Question 7. Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)

Solution :

Vertices (±2, 0), foci (±3, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the vertices are (±2, 0), = 2.

Since the foci are (±3, 0), c = 3.

We know that a2 + b2 = c2.

so, 22 + b2 = 32

b= 9 – 4 = 5

Thus, the equation of the hyperbola is.

Question 8. Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8)

Solution :

Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form .

Since the vertices are (0, ±5), = 5.

Since the foci are (0, ±8), c = 8.

We know that a2 + b2 = c2.

so, 52 + b2 = 82

b= 64 – 25 = 39

Thus, the equation of the hyperbola is

Question 9. Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5)

Solution :

Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form .

Since the vertices are (0, ±3), = 3.

Since the foci are (0, ±5), c = 5.

We know that a2 + b2 = c2.

∴32 + b2 = 52

⇒ b2 = 25 – 9 = 16

Thus, the equation of the hyperbola is .

Question 10. Foci (±5, 0), the transverse axis is of length 8.

Solution :
Given:

Foci (±5, 0), the transverse axis is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are (±5, 0), c = 5.

Since the length of the transverse axis is 8, 2a = 8 ⇒ a = 4.

We know that a2 + b2 = c2.

∴42 + b2 = 52

⇒ b2 = 25 – 16 = 9

Thus, the equation of the hyperbola is .

Question 11. Foci (0, ±13), the conjugate axis is of length 24.

Solution :
Given:

Foci (0, ±13), the conjugate axis is of length 24.

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are (0, ±13), c = 13.

Since the length of the conjugate axis is 24, 2b = 24 ⇒ b = 12.

We know that a2 + b2 = c2.

a2 + 122 = 132

⇒ a2 = 169 – 144 = 25

Thus, the equation of the hyperbola is .

Question 12. Foci the latus rectum is of length 8.

Solution :

Foci, the latus rectum is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci arec =.

Length of latus rectum = 8

We know that a2 + b2 = c2.

a2 + 4a = 45

⇒ a2 + 4a – 45 = 0

⇒ a2 + 9a – 5a – 45 = 0

⇒ (a + 9) (a – 5) = 0

⇒ a = –9, 5

Since a is non-negative, = 5.

b2 = 4= 4 × 5 = 20

Thus, the equation of the hyperbola is

Question 13. Foci (±4, 0), the latus rectum is of length 12

Solution :

Foci (±4, 0), the latus rectum is of length 12.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

We know that a2 + b2 = c2.

a2 + 6a = 16

⇒ a2 + 6a – 16 = 0

⇒ a2 + 8a – 2a – 16 = 0

⇒ (a + 8) (a – 2) = 0

⇒ a = –8, 2

Since a is non-negative, = 2.

b2 = 6= 6 × 2 = 12

Thus, the equation of the hyperbola is .

Question 14. Vertices (±7, 0),e = 4/3

Solution :

Vertices (±7, 0), e = 4/3

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the vertices are (±7, 0), = 7.

It is given that e = 4/3

We know that a2 + b2 = c2.

Thus, the equation of the hyperbola is

Question 15. Foci (0,±√10), passing through (2,3)

Solution :

Foci(0,±√10), passing through (2, 3)

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form .

Since the foci are(0,±√10), c =√10.

We know that a2 + b2 = c2.

∴ a2 + b2 = 10

⇒ b2 = 10 – a2 … (1)

Since the hyperbola passes through point (2, 3),

From equations (1) and (2), we obtain

In hyperbola, c > a, i.e., c2 > a2

∴ a2 = 5

⇒ b2 = 10 – a2 = 10 – 5 = 5

Thus, the equation of the hyperbola is .

### Miscellaneous Exercise

Question1. In 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.

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),

102 = 4a(5)

⇒ 100 = 20a

⇒ a = 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.

Question2. 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 :

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 negative y-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form x2 = – 4ay (as it is opening downwards).

It can be clearly seen that the parabola passes through point (5/2 , – 10)

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.

Question3. The cable of a uniformly loaded suspension bridge hange 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 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m 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

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 = 6 m, and.BC = 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,

∴Equation of the parabola,  or 6x2 = 625y

The x-coordinate of point D is 18.

Hence, at x = 18,

∴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 18 m from the middle is approximately 9.11 m.

Question4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Solution :

Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.

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  , where a is the semi-major axis

Accordingly, 2a = 8 ⇒ a = 4

b = 2

Therefore, the equation of the semi-ellipse is

Let A be a point on the major axis such that AB = 1.5 m.

Draw AC⊥ OB.

OA = (4 – 1.5) m = 2.5 m

The x-coordinate of point C is 2.5.

On substituting the value of x with 2.5 in equation (1), we obtain

∴AC = 1.56 m

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

Question5. A rod of length 12 cm moves with its ends 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 ΔPBQ,

In ΔPRA,

Thus, the equation of the locus of point P on the rod is .

Question6. Find the area of the triangle formed by the lines the vertex of the parabola x2 = 12to 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

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 ΔOAB are O (0, 0), A (–6, 3), and B (6, 3).

Thus, the required area of the triangle is 18 unit2.

Question7. A man running a race leave no space course notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Solution :

Let A and B be the positions of the two flag posts and P(xy) be the position of the man. Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form , where a is the semi-major axis

Accordingly, 2a = 10 ⇒ a = 5

Distance between the foci (2c) = 8

⇒ c = 4

On using the relation, we obtain

Thus, the equation of the path traced by the man is .

Question8. An equilateral triangle is inscribed n the parabola y2 = 4 ax where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution :

Let OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x-axis at point C.

Let OC = k

From the equation of the given parabola, we have

∴The respective coordinates of points A and B are

AB = CA + CB =

Since OAB is an equilateral triangle, OA2 = AB2.

Thus, the side of the equilateral triangle inscribed in parabola y2 = 4 ax is  .

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