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(a) x² + y² + 2ax + 2by + 2b² = 0 

Equation of the circle is (x – h)² + (y – k)² = r² 

⇒ (x + a)² + (y + b)² = a² – b² 

⇒ x² + y² + 2ax + 2by + a² + b² = a² – b² 

⇒ x² + y² + 2ax + 2by + 2b² = 0

(a) x + 2y + 2 = 0

Locus is line parallel to line x + 2y + 7 = 0 which is x + 2y + 2 = 0

(a) 2

Here a = 0, h = 2, b = 2 

Condition for pair of parallel lines is b² – ab = 0 

4 – a(2) = 0 

⇒ -2a = -4 

⇒ a = 2

(b) 4x – 1 = 0

y² = -x. 

It is a parabola open leftwards. 

Here 4a = 1 ⇒ a = \(\frac{1}{4}\)

Equation of directrix is x = a. 

i.e., x = \(\frac{1}{4}\) (or) 4x – 1 = 0

(b) -\(\frac{2h}{b}\)

(c) 2

Here a = k, b = -2 

Condition for perpendicular is 

a + b = 0

⇒ k – 2 = 0 

⇒ k = 2

(c) (1, -1)

To get centre we must solve the given equations 

2x – 3y – 5 = 0 … (1) 

3x – 4y – 7 = 0 … (2) 

(1) x 3 ⇒ 6x – 9y = 15 

(2) x 2 ⇒ 6x – 8y = 14 

Subtracting, -y = 1 ⇒ y = -1 

Using y = -1 in (1) we get 

2x + 3 – 5 = 0 

⇒ 2x = 2 

⇒ x = 1

(a) y² = 20x 

y² = 4(5)x

∴ a = 5

(b) x² = 8y = 4(2)y 

∴ a = 2

(c) x² = -16y = -4(4)y 

∴ a = 4

y² – 8y – 8x + 24 = 0 

⇒ y² – 8y – 4² = 8x – 24 + 4² 

⇒ (y – 4)² = 8x – 8 

⇒ (y – 4)² = 8(x – 1) 

⇒ (y – 4)² = 4(2) (x – 1) 

∴ a = 2

Y² = 4(2)X where X = x – 1 and Y = y – 4

Answer is (d) 1

y² = kx passes through (4, -2) 

(-2)² = k(4) 

⇒ 4 = 4k 

⇒ k = 1

y² = x = 4(\(\frac{1}{4}\))x 

a = \(\frac{1}{4}\)

Equation of LR is x = a or x – a = 0 

i.e., x = \(\frac{1}{4}\)

⇒ 4x = 1 

⇒ 4x – 1 = 0 

Focus (a, 0) = (\(\frac{1}{4}\), 0)

(c) (0, 4)

x² = 16y 

Here 4a = 16 ⇒ a = 4 

Focus is (0, a) = (0, 4)

(c) ±4

To get y-intercept put x = 0 in the circle equation we get 

0 + y² = 16 

∴ y = ±4

The equation of the circle when entremities (x₁, y₁) and (x₂, y₂) are given is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0 

⇒ (x – 4) (x + 2) + (y – 7) (y – 5) = 0 

⇒ x² – 2x – 8 + y² – 12y + 35 = 0 

⇒ x² + y2 – 2x – 12y + 27 = 0

Let the required of the circle be x² + y² + 2gx + 2fy + c = 0 … (1) 

It passes through (0, 1) 

0 + 1 + 2g(0) + 2f(1) + c = 0 

1 + 2f + c = 0 

2f + c = -1 … (2) 

Again the circle (1) passes through (4, 3) 

4² + 3² + 2g(4) + 2f(3) + c = 0 

16 + 9 + 8g + 6f + c = 0 

8g + 6f + c = -25 … (3) 

Again the circle (1) passes through (1, -1)

1² + (-1)² + 2g(1) + 2f(-1) + c = 0 

1 + 1 + 2g – 2f + c = 0 

2g – 2f + c = -2 … (4) 

8g + 6f + c = -25 

(4) x 4 subtracting we get, 8g – 8f + 4c = -8 

14f – 3c = -17 … (5)

14f – 3c = -17 

(2) x 3 ⇒ 6f + 3c = -3 

Adding we get 20f = -20 

f = -1 

Using f = -1 in (2) we get, 2(-1) + c = -1 

c = -1 + 2 

c = 1 

Using f = -1, c = 1 in (3) we get 

8g + 6(-1) + 1 = -25 

8g – 6 + 1 = -25 

8g – 5 = -25 

8g = -20

g = \(\frac{-20}{8}\) = \(\frac{-5}{2}\)

Using g = \(\frac{-5}{2}\), f = -1, c = 1 in (1) we get the equation of the circle. 

x² + y² + 2(\(\frac{-5}{2}\))x + 2(-1)y + 1 = 0 

x² + y² – 5x – 2y + 1 = 0

Circumference, 2πr = 16π 

⇒ 2r = 16 

⇒ r = 8 

Equation of the circle when centre and radius are known is (x – h)² + (y – k)² = r² 

⇒ (x + 3)² + (y + 2)² = 8² 

⇒ x² + 6x + 9 + y² + 4y + 4 = 64

⇒ x² + y² + 6x + 4y + 13 = 64 

⇒ x² + y² + 6x + 4y – 51 = 0

(c) \(\frac{1}{3}\)

To get x-intercept put y = 0 in 3x + 2y – 1 = 0 we get 

3x – 1 = 0 

x = \(\frac{1}{3}\)

y = -x² + 10x – 15 

⇒ y = -[x² – 10x + 5² – 5² + 15] 

⇒ y = -[(x – 5)² – 10] 

⇒ y = 10 – (x – 5)²

⇒ (x – 5)² = -(y – 10) 

This is a parabola which is open downwards.

Vertex is the maximum point. 

∴ Profit is maximum when x – 5 = 0 (or) x = 5 months. 

After that profit gradually reduces. 

∴ The best time to end the project is after 5 months.

Let output be x and average variable cost = y 

y = \(\frac{1}{5}\)x² – 6x + 100 

⇒ 5y = x² – 30x + 500 

⇒ x² – 30x + 225 = 5y – 500 + 225 

⇒ (x – 15)² = 5y – 275 

⇒ (x – 15)² = 5(y – 55) which is of the form X² = 4(\(\frac{5}{4}\))Y

∴ Y average variable cost curve is a parabola Vertex (0, 0) 

x – 15 = 0; y – 55 = 0 

x = 15; y = 55 

At the vertex, output is 15 tonnes and average cost is Rs 55.

The condition for a line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a² (1 + m²) 

Equation of the line is 3x + 4y – P = 0 

4y = -3x + P 

y = \(\frac{-3}{4}x+\frac{P}{4}\)

∴ m = \(\frac{-3}{4}\), c = \(\frac{P}{4}\)

Equation of the circle is x² + y² = 16 

∴ a² = 16

Condition for tangency we have c² = a²(1 + m²)

⇒ \((\frac{P}{4})^2\) =  16(1 + \(\frac{9}{16}\))

⇒ \(\frac{P^2}{16}\) = 16(\(\frac{25}{16}\))

⇒ P² = 16 x 25 

⇒ P = ±√16√25 

⇒ P = ±4 x 5 

⇒ P = ±20

The given equation is 4x² + 12xy + 9y² – 6x – 9y + 2 = 0 

Here a = 4, 2h = 12, (or) h = 6 and b = 9 

h² – ab = 6² – 4 x 9 = 36 – 36 = 0 

∴ The given equation represents a pair of parallel straight lines 

Consider 4x² + 12xy + 9y² = (2x)² + 12xy + (3y)² 

= (2x)² + 2(2x)(3y) + (3y)² 

= (2x + 3y)² 

Here we have repeated factors. 

Now consider, 4x² + 12xy + 9y² – 6x – 9y + 2 = 0 

(2x + 3y)² – 3(2x + 3y) + 2 = 0 

t² – 3t + 2 = 0 where t = 2x + 3y 

(t – 1)(t – 2) = 0 

(2x + 3y – 1) (2x + 3y – 2) = 0

∴ Separate equations are 2x + 3y – 1 = 0, 2x + 3y – 2 = 0

(i) Equation of the circle is (x – h)² + (y – k)² = r² 

Centre (h, k) = (3, 5) and radius r = 5 

∴ Equation of the circle is (x – 3)² + (y – 5)² = 5² 

⇒ x² – 6x + 9 + y² – 10y + 25 = 25 

⇒ x² + y² – 6x – 10y + 9 = 0

(ii) Equation of the circle when centre origin (0, 0) and radius r is x² + y² = r² 

⇒ x² + y² = 2² 

⇒ x² + y² = 4 

⇒ x² + y² – 4 = 0

The equation of the tangent to the circle x² + y² – 4x + 4y – 8 = 0 at (x₁, y₁) is 

xx₁ + yy₁ – 4\(\frac{(x+x_1)}{2}\) + 4\(\frac{(x+y_1)}{2}\) – 8 = 0 

Here (x₁, y₁) = (-2, -2) 

⇒ x(-2) + y(-2) – 2(x – 2) + 2(y – 2) – 8 = 0 

⇒ -2x – 2y – 2x + 4 + 2y – 4 – 8 = 0

⇒ -4x – 8 = 0 

⇒ x + 2 = 0

The equation of the circle is x² + y² – 4x – 6y + 9 = 0

PT² = x₁² + y₁² – 4x₁ – 6y₁ + 9 

At P(1, 0), PT² = 1 + 0 – 4 – 0 + 9 = 6 > 0 

At Q(2, 1), PT² = 4 + 1 – 8 – 6 + 9 = 0 

At R(2, 3), PT² = 4 + 9 – 8 – 18 + 9 = -4 < 0 

The point P lies outside the circle. 

The point Q lies on the circle. 

The point R lies inside the circle.

(b) \(-\frac{7}{5}\)

Slope of 7x + 5y – 8 = 0 is = \(\frac{-x\,coefficient}{y\,coefficient}\) = \(-\frac{7}{5}\)

(a) 25

y² = -25a

Here 4a = 25 which is the length of the latus rectum.

Answer is (b) 2a

Let the point P(x₁, y₁). 

Fixed points are A(-1, 1) and B(2, 3). 

Given area (formed by these points) of the triangle 

APB = 8

⇒ \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 8 

⇒ \(\frac{1}{2}\)[x₁(1 – 3) + (-1) (3 – y₁) + 2(y₁ – 1)] = 8 

⇒ \(\frac{1}{2}\)[-2x₁ – 3 + y₁ + 2y₁ – 2] = 8 

⇒ \(\frac{1}{2}\)[-2x₁ + 3y₁ – 5] = 8 

⇒ -2x₁ + 3y₁ – 5 = 16 

⇒ -2x₁ + 3y₁ – 21 = 0 

⇒ 2x₁ – 3y₁ + 21 = 0 

∴ The locus of the point P(x₁, y₁) is 2x – 3y + 21 = 0.