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Given equation of the ellipse is 16x² + 25y² = 400

\(= \frac{x^2}{25}+\frac{y^2}{16}=1\\=\frac{x^2}{5^2}+\frac{y^2}{4^2}=1\) 

a = 5,b = 4

∴ Length of major axis = 2a = 10 

And length of minor axis = 2b = 8.

Correct option is: (B) y = ±4

Let, the equation of the circle with centre (h, k) and radius ‘r’ be

(x − h)² + (y − k)² = r²  … (1) 

Since the circle passes through (2, −2) and (3, 4). 

(2 − h)² + (−2 − k)² = r²  … (2) 

And (3 − h)² + (4 − k)² = r² … (3) 

(2) and (3) ⇒ (2 − h)² + (2 + k)² = (3 − h)² + (4 − k)²

⇒ 4 − 4h + h² + 4 + 4k + k² = 9 − 6h + h² + 16 − 8k + k² 

⇒ −4h + 4k+ 8 + 6h + 8k − 25 = 0 

⇒ 2h + 12k − 17 = 0  … (4) 

Also, centre (h, k) lies on x + y = 2 

∴h + k = 2  … (5) 

(4) − 2 × (5) ⇒ 10k − 17 = −4

⇒ \(k=\frac{13}{10}\)

(5) ⇒ k = 1.3 ⇒ h= 0.7 

∴ (1) ⇒ (x − 0.7)² + (y − 1.3)² = r² … (6)

(2) ⇒ (2 − 0.7)² + (−2 − 1.3)² = r² 

⇒ r² = 12.58 ∴ (6) 

⇒ (x − 0.7)²+ (y − 13)² = 12.58, is the required equation of the circle.

Answer is (B) y = 2

Equation of parabola

x² = – 8y

⇒ x² = -4 × 2 × y

Here a = 2

By equation of directrix y = a

y = 2

Answer is (C) e > 0

Answer is (C) (2, 2)

Equation of line,

2y – x = 2

x = 2y – 2 ….(i)

Equation of parabola,

y² = 2x

From equation (i) and (ii),

y² = 2(2y – 2)

⇒ y² = 4y – 4

⇒ y² – 4y + 4 = 0

⇒ (y – 2)² = 0

⇒ y – 2 = 0

⇒ y = 2

Put value of y in equation (i),

x = 2 × 2 – 2 = 2

Thus tangent point, (x, y) = (2, 2)

Centre of circle = (r, h)

Radius of circle = r

Thus, equation of circle

(x – r)² + (y – h)² = r²

Let tangent OB touches the circle at B. 

Tangent passes through origin. 

Let tangent touches the circle at point (x₁,y₁).

∴ Equation of tangent at point of circle,

xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0

xx₁ + yy₁ – r(x + x₁) + h(y + y₁) + h² = 0

Since the line passes through origin.

∴ x × 0 + y × 0 – r(x + 0) – h(y + 0) + h² = 0

⇒ – rx – hy + h² = 0

⇒ rx + hy – h² = 0

This is required equation.

From point (x₁, y₁) two tangents can be drawn at parabola whose combined equation can be find by equation SS’ = T²

Given Point – (4, 0)

Equation of parabola,

⇒ y² = 8x where a = 2

then S = y² – 4ax₁

⇒ S = y² – 8x …(i)

S’ = y1² – 4ax²

⇒ S’ = (10)² – 4 × 2 × 4

⇒ S’ = 100 – 32 = 68 …(ii)

T = yy1 – 2a(x + x₁)

⇒ T = y × 10 – 2 × 2 × (x + 4)
= T = 10y – 4(x + 4)

T² = {10y – 4(x + 4)}²

⇒ T² = 100 y² + 16(x + 4)² – 80(x + 4)y

⇒ T² = 100y² + 16x² + 256 + 128x – 80xy- 320y …(iii)

Put the values from eqn. (i), (ii), (iii) in SS’ – T²

(y² – 8x) (68) = 16x² + 100y² – 80xy + 128x – 320y + 256

⇒ 68y² – 744x – 16x² – 100y²
+ 80xy – 128x+ 320y – 256 = 0

⇒ – 16x² – 32y² + 80xy – 672x + 320y – 256 = 0

⇒ x² + 2y² – 5xy + 42x – 20y + 16 = 0

This is required equation.

Answer is  (C) c = ±√ (a²m² + b²)

Equation of line, x – 3y – 4 =0

⇒ x = 3y + 4

Equation of ellipse,

3x² + 4y² = 20

Putting value of x from (i) in eqn. (ii),

3(3y + 4)² + 4y² = 20

⇒ 3(9y² + 24y + 16) + 4y² – 20 = 0

⇒ 27y² + 72y + 48 + 4y² – 20 = 0

⇒ 31y² + 72y + 28 = 0

Line will touch eqⁿ. (ii) if

(72)² – 4(31 )(28) = 0

∴ (72)² – 4 × 31 × 28 = 5184 – 3472

= 1712 ≠ 0

Thus, given line will not be touch given ellipse.

∴ Given equation is wrong.

Answer is (D) y + x = 1

Equation of parabola

y² = 4x

y² = 4 × 1 × x

a = 1

Given equation of the parabola is y² = 4x 

Comparing this equation with y² = 4ax, we get 

⇒ 4a = 4 

⇒ a = 1 

Let the equation of common tangent be 

y = mx + 1/m …..(i)

Substituting y = mx + 1/m in x² = 32y, we get

⇒ x² = 32(mx + 1/m) = 32 mx + 32/m

⇒ mx² = 32 m² x + 32 

⇒ mx² – 32 m² x – 32 = 0 ……..(ii)

Line (i) touches the parabola x² = 32y. 

The quadratic equation (ii) in x has equal roots. 

Discriminant = 0 

⇒ (-32m²)² – 4(m)(-32) = 0 

⇒ 1024 m⁴ + 128m = 0 

⇒ 128m (8m³ + 1) = 0 

⇒ 8m³ + 1 = 0 …..[∵ m ≠ 0] 

⇒ m³ = - 1/8

⇒ m = - 1/2

Substituting m = - 1/2  in (i), we get

⇒ y = - 1/2 x + 1/ (- 1/2)

⇒ y = - 1/2 x-2

⇒ x + 2y + 4 = 0, which is the equation of the common tangent.

Given equation of the parabola is y² = 18x 

Comparing this equation with y² = 4ax, we get 

⇒ 4a = 18

⇒ a = 9/2

Equation of tangent to the parabola y² = 4ax having slope m is

⇒ y = mx + a/m

⇒ y = mx + 9/2m

⇒ 2ym = 2xm² + 9 

⇒ 2xm² – 2ym + 9 = 0

The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.

m₁ + m₂ = (-2y)/2x = y/x

But, m₁ + m₂ = -3

y/x = -3

y = -3x, which is the required equation of locus

Given equation of the parabola is y² = 36x. 

Comparing this equation with y² = 4ax, we get 

⇒ 4a = 36 

⇒ a = 9 

Equation of tangent to the parabola y² = 4ax having slope m is 

y = mx + \(\frac {a}{m}\)

Since the tangent passes through the point (2, 9), 

⇒ 9 = 2m + 9/m

⇒ 9m = 2m² + 9 

⇒ 2m² – 9m + 9 = 0 

⇒ 2m² – 6m – 3m + 9 = 0 

⇒ 2m(m – 3) – 3(m – 3) = 0 

⇒ (m – 3)(2m – 3) = 0 

⇒ m = 3 or m = 3/2

These are the slopes of the required tangents. 

By slope point form, y – y₁ = m(x – x₁), the equations of the tangents are 

⇒ y – 9 = 3(x – 2) and y – 9 = 3/2 (x – 2)

⇒ y – 9 = 3x – 6 and 2y – 18 = 3x – 6 

⇒ 3x – y + 3 = 0 and 3x – 2y + 12 = 0

Given equation of the parabola is y² = kx 

Comparing this equation with y² = 4ax, we get 

⇒ 4a = k 

⇒ a = k/4

Equation of tangent to the parabola y² = 4ax having slope m is

y = mx + \(\frac {a}{m}\)

Since the tangent passes through the point (-6, 9), 

⇒ 9 = -6m + k/4m

⇒ 36m = -24m2 + k

⇒ 24m² + 36m – k = 0 

The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents. 

m₁ m₂ = -k/24

Since the tangents are perpendicular to each other

m₁ m₂ = -1 

⇒ -k/24 = -1

⇒ k = 24

Alternate method: 

We know that, tangents drawn from a point on directrix are perpendicular. 

(-6, 9) lies on the directrix x = -a. 

⇒ -6 = -a 

⇒ a = 6 

Since 4a = k 

⇒ k = 4(6) = 24

Given equation of the parabola is 

3y² = 8x

\(=y^2=\frac{8}{3}x\\=y^2=4.\frac{2}{3}x\)

Here, a = \(\frac{2}{3}\)

∴ focus = (a, 0) = ( \(\frac{2}{3}\) , 0)

And length of latus rectum = 4a = \(\frac{8}{3}\)

(i) According to question,

Centre of circle (h, k) = (- 2, 3)

and Radius of circle r = 4 unit

Then, equation of circle

From formula (x – h)² + (y- k)² = r²

[x – (- 2)]² + (y – 3)² = 42

⇒ (x + 2)² + (y – 3)² = 16

⇒ x² + 4x + 4 + y² – 6y + 9 = 16

⇒ x² + y² + 4x – 6y = 16 – 9 – 4

⇒ x² + y² + 4x – 6y = 3

⇒ x² + y² + 4x – 6y – 3 = 0

Thus, required equation of circle

x² + y² + 4x – 6y – 3 = 0.

(ii) According to question,

Centre of circle (h, k) = (a, b)

and Radius of circle r = (a – b) unit

Then, equation of circle
From formula (x – h)² + (y – k)² = r²

⇒ (x – a)² + (y – b)² = (a – b)²

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

⇒ x² + y² – 2ax – 2by + a² + b² – a² – b² + 2ab = 0

⇒ x² + y² – ax – 2by + 2ab = 0.

Given equation of the parabola is y = x² – 2x + 3 

⇒ y = x – 2x + 1 + 2

⇒ y – 2 = (x – 1)² 

⇒ (x – 1)² = y – 2 

Comparing this equation with X² = 4bY, we get 

X = x – 1, Y = y – 2 

⇒ 4b = 1 

⇒ b = 1/4

The co-ordinates of vertex are (X = 0, Y = 0) 

⇒ x – 1 = 0 and y – 2 = 0 

⇒ x = 1 and y = 2 

The co-ordinates of vertex are (1, 2). 

The co-ordinates of focus are S(X = 0, Y = b) 

⇒ x – 1 = 0 and y – 2 = 1/4

⇒ x = 1 and y = 9/4

The co-ordinates of focus are (1, 9/4)

Equation of the axis is X = 0 

x – 1 = 0, i.e., x = 1 

Equation of directrix is Y + b = 0 

⇒ y – 2 + 1/4 =0

⇒ y – 7/4 = 0

⇒ 4y – 7 = 0