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Given equation of the ellipse is \(\frac {x^2}{25} + \frac {y^2}{9}=1\) 

Comparing this equation with \(\frac {x^2}{a^2} + \frac {y^2}{b^2}=1\)

we get 

a² = 25 and b² = 9 

a = 5 and b = 3

Since a > b, 

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10 

Length of minor axis = 2b = 2(3) = 6 

Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac {\sqrt{a^2-b^2}}{a}\)

= \(\frac {\sqrt{25-9}}{5}\)

= \(\frac {\sqrt16}{5}\)

= 4/5

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0), i.e., S(5(4/5), 0) and S'(-5(4/5), 0)

i.e., S(5(4, 0) and S' (-4, 0)

(iii) Equations of the directrices are x = ± a/e

= ± 5/4

= ± 25/4

(iv) Length of latus rectum = \(\frac {2b^2}{a} = \frac {2(3)^2}{5} = \frac {18}{5}\)

(v) Distance between foci = 2ae

= 2(5) (4/5)

= 8

(vi) Distance between directrices = 2a/e

= \(\frac {2(5)}{\frac{4}{5}}\)

= 25/2

Let the required equation of hyperbola be \(\frac{x^2}{a^2} - \frac {y^2}{b^2} = 1\)……(i)

Given, eccentricity (e) = 3/2

Co-ordinates of foci are (±ae, 0). 

Given co-ordinates of foci are (±2, 0) ae = 2

⇒ a(3/2) =2

⇒ a = 4/3

⇒ a² = 16/9

Let the required equation of hyperbola be \(\frac{x^2}{a^2} - \frac {y^2}{b^2} = 1\)

Length of transverse axis = 2a 

Given, length of transverse axis = 8 

⇒ 2a = 8 

⇒ a = 4 

⇒ a² = 16 

Distance between foci = 2ae 

Given, distance between foci = 10 

⇒ 2ae = 10 

⇒ ae = 5

⇒ a² e² = 25 

Now, b² = a² (e² – 1) 

⇒ b² = a² e² – a² 

⇒ b² = 25 – 16 = 9

The required equation of hyperbola is \(\frac {x^2}{16} - \frac {y^2}{9}=1\)

Let the required equation of hyperbola be \(\frac{x^2}{a^2} - \frac {y^2}{b^2} = 1\)

Length of transverse axis = 2a 

Given, length of transverse axis = 6 

⇒ 2a = 6 

⇒ a = 3 

⇒ a² = 9 

Length of conjugate axis = 2b 

Given, length of conjugate axis = 9

⇒ 2b = 9

⇒ b = 9/2

⇒ b² = 81/4

The required equation of hyperbola is \(\frac {x^2}{9} - \frac {y^2}{\frac{81}{4}}=1\)

i.e., \(\frac {x^2}{9} - \frac {4y^2}{81}=1\)

 Given, one of the foci of the hyperbola is  x² – 3y² = 3

\(\frac{x^2}{3} - \frac {y^2}{1} = 1\)

Equation of the hyperbola conjugate to the above hyperbola is \(\frac {y^2}{1}-\frac{x^2}{3} = 1\)

Comparing this equation with \(\frac{x^2}{b^2} - \frac {y^2}{a^2} = 1\)

we get b² = 1 and a² = 3 

Now, a² = b² (e² – 1) 

⇒ 3 = 1(e² – 1) 

⇒ 3 = e – 1 

⇒ e² = 4 

⇒ e = 2 …..[∵ e > 1]

Given the equation of the parabola is y² = 8x. 

Comparing this equation with y² = 4ax, we get 

4a = 8 

a = 2 

Slope of the line 2x + 2y + 5 = 0 is -1 

Since the tangent is parallel to the given line, slope of the tangent line is m = -1 

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

y = mx + a/m 

Equation of the tangent is

y = -x + 2/-1

x + y + 2 = 0

Point of contact =(\(\frac{a}{m^2}, \frac{2a}{m}\))

= \(\frac {2}{(-1)^2}, \frac{2(2)}{-1}\)

= (2, -4)

Given equation of the parabola is y² = 8x 

Comparing this equation with y² = 4ax, we get 

4a = 8 

a = 2 

t = 1 

Equation of tangent with parameter t is yt = x + at² 

∴ The equation of tangent with t = 1 is 

y(1) = x + 2(1)² 

y = x + 2 

∴ x – y + 2 = 0

Equation of parabola,

x² – 4x – 8y = 4

⇒ x² – 2.2x + (2)² = 8y + 4 + 2²

⇒ (x – 2)² = 8y + 8

= 8(y + 1)

(x-2)² = 4.2.(y + 1)

x² = 4.2.y

Where x = x – 2 and Y = y + 1

a = 2

Coordinates of focus = (0, a)

x = x- 2 = 0

⇒ x = 2

Y = y + 1 = 2

⇒ y = 1

Thus coordinates of focus = (2, 1)

Given equation of the parabola is 3x² = 8y 

⇒ x² = 8/3 y

Comparing this equation with x² = 4by, we get 

⇒ 4b = 8/3

⇒ b = 2/3

Co-ordinates of focus are S(0, b), i.e., S(0, 2/3) 

Equation of the directrix is y + b = 0, 

⇒ y + 2/3 = 0

⇒ 3y + 2 = 0 

Length of latus rectum = 4b = 4 (2/3) = 8/3

Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b),

⇒ (4/3, 2/3) and (- 4/3, 2/3).

The equation of the given circle is x² + y² – 4x – 8y – 45 = 0.
x² + y² – 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
⇒ ( – 2)² + ( –4)²= √65 ,
Which is of the form (x – h)² + (y – k)² = r², where h = 2, k = 4 and = √65 ,
Thus, the centre of the given circle is (2, 4), while its radius is √65 .

The equation of the given circle is x² + y² – 8x + 10y – 12 = 0.
x² + y² – 8x + 10y – 12 = 0
⇒ (x² – 8x) + (y² + 10y) = 12
⇒ {x² – 2(x)(4) + 4²} + {y² + 2(y)(5) + 5²}– 16 – 25 = 12
⇒ (x – 4)² + (y + 5)2 = 5³

 ⟹(−4)²+{−(−5)}²=(√53)²
which is of the form (x – h)² + (y – k)² = r², where h = 4, k = – 5 and r = √53
Thus, the centre of the given circle is (4, –5), while its radius is √53.

Given equation of the ellipse is \(\frac {x^2}{25} + \frac {y^2}{9} = 1\)

Comparing this equation with \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\),

we get a² = 25 and b² = 9 

∴ a = 5 and b = 3

Since a > b, 

X-axis is the major axis and Y-axis is the minor axis. 

(i) Length of major axis = 2a = 2(5) = 10 

Length of minor axis = 2b = 2(3) = 6 

∴ Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^2-b^2}}{a}\)

∴ e =\(\frac{\sqrt{25-9}}{5}\)= 4/5

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., S(5(4/5), 0) and S'(-5(4/5), 0)

i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ± a/e

i.e., x = ± \(\frac {5}{\frac{4}{5}}\)

i.e., x = ± 25/4

(iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(3)^2}{5} = \frac {18}{5}\)

(v) Distance between foci = 2ae = 2 (5) (4/5) = 8

(vi) Distance between directrices = 2a/e = \(\frac {2(5)}{\frac{4}{5}}\)= 25/2

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

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

Comparing this equation with \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\),

we get a² = 25 and b² = 16

∴ a = 5 and b = 4

Since a > b, 

X-axis is the major axis and Y-axis is the minor axis. 

(i) Length of major axis = 2a = 2(5) = 10 

Length of minor axis = 2b = 2(4) = 8

∴ Lengths of the principal axes are 10 and 8.

(ii) b² = a² (1 – e²)

16 = 25(1 – e²)

16/25 = 1- e²

e² = 1- 16/25

e² = 1- 9/25

e = 3/5……[∵ 0 < e < 1]

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., S(5(3/5), 0) and S'(-5(3/5), 0)

i.e., S(3, 0) and S'(-3, 0)

(iii) Equations of the directrices are x = ± a/e

i.e., x = ± \(\frac {5}{\frac{3}{5}}\)

i.e., x = ± 25/3

(iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(16)}{5} = \frac {32}{5}\)

(v) Distance between foci = 2ae = 2 (5) (3/5) = 6

(vi) Distance between directrices = 2a/e = \(\frac {2(5)}{\frac{3}{5}}\)= 50/3

Given equation of the ellipse is \(\frac {x^2}{144} - \frac {y^2}{25} = 1\)

Comparing this equation with \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\),

we get a² = 144 and b² = 25

∴ a = 12 and b = 5

(i) Length of major axis = 2a = 2(12) = 24

Length of minor axis = 2b = 2(5) = 10

∴ Lengths of the principal axes are 24 and 10.

(ii) b² = a² (e² – 1)

25 = 144 (e² – 1)

25/144 = e² – 1

e² = 1 + 25/144

e² = 169/144

e = 13/12 …….[∵ e > 1]

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., S(12(13/12), 0) and S'(-12(13/12), 0)

i.e., S(13, 0) and S'(-13, 0)

(iii) Equations of the directrices are x = ± a/e

i.e., x = ± \(\frac {12}{\frac{13}{12}}\)

i.e., x = ± 144/13

(iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(25)}{12} = \frac {25}{6}\)

(v) Distance between foci = 2ae = 2 (12) (13/12) = 26

(vi) Distance between directrices = 2a/e = \(\frac {2(12)}{\frac{13}{12}}\)= 288/13

Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0). The foci of the ellipse are on the X-axis.

Let the required equation of ellipse be \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) where a > b.

Length of latus rectum = 2b²/a

2b²/a =6

b² = 3a …..(i) 

Co-ordinates of foci are (±ae, 0) 

ae = 2

a² e² = 4 …..(ii)

Now, b² = a² (1 – e²)

b² = a² – a² e² 

3a = a² – 4 …..[From (i) and (ii)] 

a² – 3a – 4 = 0 

a² – 4a + a – 4 = 0 

a(a – 4) + 1(a – 4) = 0 

(a – 4) (a + 1) = 0

a – 4 = 0 or a + 1 = 0 

a = 4 or a = -1 

Since a = -1 is not possible, 

a = 4

a² = 4

Substituting a = 4 in (i), we get b = 3(4) = 12

The required equation of ellipse is \(\frac {x^2}{16}+\frac{y^2}{12}=1\)

 Given equation of the ellipse is x² – y² = 16

Comparing this equation with \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\),

we get a² = 16 and b² = 16

∴ a = 4 and b = 4

(i) Length of major axis = 2a = 2(4) = 8

Length of minor axis = 2b = 2(4) = 8

(ii) We know that e = \(\frac{\sqrt{a^2-b^2}}{a}\)

∴ e =\(\frac{\sqrt{16+16}}{4}\)

= √32/4

= 4√2/4

= √2

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., S(4√2), 0) and S'(-4√2, 0)

(iii) Equations of the directrices are x = ± a/e

i.e., x = ± 4/√2

i.e., x = ± 2√2

(iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(16)}{4} = 8\)

(v) Distance between foci = 2ae = 2 (4) (√2) = 8√2

(vi) Distance between directrices = 2a/e = \(\frac {2(4)}{\sqrt2}\)= 4√2

Correct option is: (B) x² + 4y² = 100 

Correct option is:(C) 128x² + 144y² = 18432 

we write the given equation in the form (x² – 2x) + ( y² + 4y) = 8

Now, completing the squares, we get

(x² – 2x + 1) + ( y₂ + 4y + 4) = 8 + 1 + 4

(x – 1)² + (y + 2)² = 13

Comparing it with the standard form of the equation of the circle, we see that the centre of the circle is (1, –2) and radius is √13

The given equation is x² = –9y.
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x² = –4ay, we obtain

- 4a = -9 ⇒ b = 9/4

∴Coordinates of the focus = (0, a ) =(0, - 9/4)

Since the given equation involves x2, the axis of the parabola is the y-axis.
Equation of directrix, y = a,i.c., y =9/4
Length of latus rectum = 4a = 9

The given equation is y² = 10x.
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax, we obtain

4a =10 ⇒a =5/2

∴Coordinates of the focus = (a, 0)=(5/2,0)

Since the given equation involves y2, the axis of the parabola is the x-axis.
Equation of directrix, x = -a,i.c.,x = -5/2
Length of latus rectum = 4a = 10

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

Focus of the parabola is S(-7, 0) and vertex is O(0, 0). 

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

Focus S(-7, 0) lies on the left-hand side of the origin. 

It is a left-handed parabola. Required parabola is y = -4ax. 

Focus is S(-a, 0). a = 7 

∴ The required equation of the parabola is y² 

=-4(7)x, i.e., y² = -28x.

The distance of any point on the parabola from its focus and its directrix is same.

Given that, directrix, x = 0 and focus = (6, 0)

If a parabola has a vertical axis, the standard form of the equation of the parabola is (x - h)² = 4p(y - k), where p≠ 0.

The vertex of this parabola is at (h, k).

The focus is at (h, k + p) & the directrix is the line y = k - p.

As the focus lies on x – axis,

Equation is y² = 4ax or y² = -4ax

So, for any point P(x, y) on the parabola

Distance of point from directrix = Distance of point from focus

x² = (x – 6)² + y²

x² = x² - 12x + 36 + y²

y² - 12x + 36 = 0

Hence the required equation is y² - 12x + 36 = 0.

Solving the given equations,

3x + y = 14 ……….1

2x + 5y = 18 …………..2

Multiplying the first equation by 5, we get

15x + 5y = 70…….3

2x + 5y = 18……..4

Subtract equation 4 from 3 we get

13 x = 52,

Therefore x = 4

Substituting x = 4, in equation 1, we get

3 (4) + y = 14

y = 14 – 12 = 2

So, the point of intersection is (4, 2)

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

Putting the values of (4, 2) and centre co-ordinates (1,-2) in the above expression, we get

(4 – 1)² + (2 – (-2))² = r²

3² + 4² = r²

r² = 9 + 16 = 25

r = 5 units

So, the expression is

(x – 1)² + (y – (-2))² = 5²

Expanding the above equation we get

x² – 2x + 1 + (y + 2)² = 25

x² – 2x + 1 + y² + 4y + 4 = 25

x² – 2x + y² + 4y – 20 = 0

Hence the required expression is x² – 2x + y² + 4y – 20 = 0.

When circle touches both the axes, the co – ordinates of the centre and its radius are equal in their magnitude,

h = k – r

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(3 – h)² + (6 – h)² = h²

9 + h² - 6h + 36 + h² - 12h = h²

h² - 18h + 45 = 0

h² - 15h – 3h + 45 = 0

h (h – 15) – 3 (h – 15) = 0

(h – 3) (h – 15) = 0

h = 3 or h = 15

Co – ordinates of centre are (3, 3) or (15, 15)

(x – h)² + (y – k)² = r²

Equation, having centre (3, 3)

(x – 3)² + (y – 3)² = 3²

x² - 6x + 9 + y² - 6y + 9 – 9 = 0

x² - 6x + y² - 6y + 9 = 0

Equation, having centre (15, 15)

(x – 15)² + (y – 15)² = 15²

x² - 30x + 225 + y² - 30y + 225 – 225 = 0

x² - 30x + y² - 30y + 225 = 0

Hence the equations are x² - 6x + y² - 6y + 9 = 0 or x² - 30x + y² - 30y + 225 = 0.

Option (C) is the answer.

Given equation of the circle,

x² – 4x + y² – 6y + 11 = 0

x² – 4x + 4 + y² – 6y + 9 +11 – 13 = 0

the above equation can be written as

x² – 2 (2) x + 2² + y² – 2 (3) y + 3² +11 – 13 = 0

on simplifying we get

(x – 2)² + (y – 3)² = 2

(x – 2)² + (y – 3)² = (√2)²

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

We have centre = (2, 3)

The centre point is the mid-point of the two ends of the diameter of a circle.

Let the points be (p, q)

(p + 3)/2 = 2 and (q + 4)/2 = 3

p + 3 = 4 & q + 4 = 6

p = 1 & q = 2

Hence, the other ends of the diameter are (1, 2).

The correct option is (A). The point of intersection of 3x + y – 14 = 0 and 2x + 5y – 18 = 0 are x = 4, y = 2, i.e., the point (4, 2)

Therefore, the radius is = √(9+16)=5 and hence the equation of the circle is given by

(x – 1)² + (y + 2)² = 25

or, x² + y² – 2x + 4y – 20 = 0.

The correct choice is (A), since the equation can be written as (x – 1)² + (y – 1)² = 1 which represents a circle touching both the axes with its centre (1, 1) and radius one unit.

Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.

Correct option is:(A) (±√13, 0)

Given equation of the parabola is 2y² = 7x.

 ⇒ y² = 7/2 x

Comparing this equation with y² = 4ax, we get

⇒ 4a = = 7/2

⇒ a = = 7/8

If t is the parameter of the point P on the parabola, then 

P(t) = (at² , 2at) 

i.e., x = at² and y = 2at …..(i)

Given, t = -2 

Substituting a = 7/8 and t = -2 in (i), we get

x = 7/8 (-2)2 and y = 2 (7/8) (-2)

x = 7/2 and y = -7/2

The co-ordinates of the point on the parabola are

(7/2, -7/2)

∴ Focal distance = x + a

= 7/2 + 7/8

= 35/8

Given equation of the parabola is y² = 12x. Comparing this equation with y² = 4ax, we get 

⇒ 4a = 12 

⇒ a = 3 

If t is the parameter of the point P on the parabola, then 

P(t) = (at² , 2at) 

i.e., x = at² and y = 2at ……..(i)

Given, t = 1/3

Substituting a = 3 and t = 1/3 in (i), we get 

x = 3 (1/3)² and y = 2(3)(1/3)

x = 1/3 and y =2

The co-ordinates of the point on the parabola are (1/3, 2)

∴ Focal distance = x + a

= 1/3 + 3

= 10/3

Given the equation of the parabola is y² = 16x. 

Comparing this equation with y² = 4ax, we get 

⇒ 4a = 16 

⇒ a = 4 

Since ordinate is 2 times the abscissa, y = 2x

Substituting y = 2x in y² = 16x, we get 

⇒ (2x)² = 16x 

⇒ 4x² = 16x 

⇒ 4x² – 16x = 0 

⇒ 4x(x – 4) = 0 

⇒ x = 0 or x = 4 

When x = 4, 

focal distance = x + a = 4 + 4 = 8 

When x = 0, 

focal distance = a = 4 

∴ Focal distance is 4 or 8.

Vertex of the parabola is at origin (0, 0) and its axis is along X-axis. 

Equation of the parabola can be either y² = 4ax or y ²= -4ax. 

Since the parabola passes through (2, 3), it lies in 1st quadrant. 

∴ Required parabola is y² = 4ax. 

Substituting x = 2 and y = 3 in y² = 4ax, we get 

⇒ (3)² = 4a(2)

⇒ 9 = 8a 

⇒ a = 9/8

The required equation of the parabola is 

⇒ y² = 4 (9/8) x

⇒ y² = 9/2 x

⇒ 2y² = 9x.

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

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

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

Answer 10:
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 X²/a² - Y²/b² =1.
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 X²/16 - Y²/9 =1. 

Vertices (0, ±3), foci (0, ±5)
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form Y²/a² - X²/b² =1.
Since the vertices are (0, ±3), a = 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 Y²/9 - X²/16 =1.

4x² + 9y² = 72 & 2x + 3y = 12

Solving both the expressions,

2x = 12 – 3y

Squaring both sides,

(2x)² = (12 – 3y)²

4x² = (12 – 3y)²

Putting the value of 4x²,

(12 – 3y)² + 9y² = 72

144 + 9y² - 72y + 9y² - 72 = 0

18y² - 72y + 72 = 0

Y² - 4y + 4 = 0

(y – 2)²= 0

y = 2

2x = 12 – 3y

2x = 12 – 3(2)

2x = 12 – 6 = 6

x = 3

So, the point of intersection is (3,2).

TRUE

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

Let the required equation of hyperbola be \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\)

Length of conjugate axis = 2b 

Given, length of conjugate axis = 5 

2b = 5

b = 5/2

b² = 25/4

Distance between foci = 2ae 

Given, distance between foci = 13 

2ae = 13

ae = 13/2

a² e² = 169/4

Now, b² = a² (e² – 1)

b² = a² e² – a²

25/4 = 169/4 - a²

a² = 169/4 - 25/4 = 36

∴ The required equation of hyperbola is \(\frac {x^2}{36} - \frac {y^2}{\frac {25}{4}} = 1\)

i.e., \(\frac {x^2}{36} - \frac {4y^2}{25} = 1\)

Given equation of the parabola is y² = 12x 

Comparing this equation with y² = 4ax, we get 4a = 12 

∴ a = 3 

If t is the parameter of the point P on the parabola, then 

P(t) = (at² , 2at)

i.e., x = at² and y = 2at …..(i) 

(i) Given, t = 2 

Substituting a = 3 and t = 2 in (i), we get 

x = 3(2)² and y = 2(3)(2) 

x = 12 and y = 12 

∴ The cartesian co-ordinates of the point on the parabola are (12, 12)

(ii) Given, t = -3

Substituting a = 3 and t = -3 in (i), we get 

x = 3(-3)² and y = 2(3)(-3) 

x = 27 and y = 18

∴ The cartesian co-ordinates of the point on the parabola are (27, 18)

Given equation of the parabola is y² = 12x. Comparing this equation with y² = 4ax, we get 

⇒ 4a = 12

⇒ a = 3 

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, 5) 

⇒ 5 = 2m + 3/m

⇒ 5m = 2m² + 3 

⇒ 2m² – 5m + 3 = 0 

⇒ 2m² – 2m – 3m + 3 = 0 

⇒ 2m(m – 1) – 3(m – 1) = 0 

⇒ (m- 1)(2m – 3) = 0

⇒ m = 1 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 – 5 = 1(x – 2) and y – 5 = 3/2 (x – 2) 

⇒ y – 5 = x – 2 and 2y – 10 = 3x – 6 

⇒ x – y + 3 = 0 and 3x – 2y + 4 = 0

Circle, parabola, ellipse and hyperbola.

Vertex of the parabola is at origin (0, 0) and its axis is along Y-axis.

Equation of the parabola can be either x = 4by or x² = -4by 

Since the parabola passes through (-10, -5), it lies in 3rd quadrant. 

Required parabola is x² = -4by. 

Substituting x = -10 and y = -5 in x² = -4by, we get 

⇒ (-10)² = -4b(-5)

⇒ b = 100/5 = 5

∴ The required equation of the parabola is x² = -4(5)y, i.e., x² = -20y.

An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called foci. 

Points to be noted: 

• The mid point of the line segment joining the foci is called centre of the ellipse. 
• The line segment through the foci is called the major axis. 
• The line segment through the centre and perpendicular to the major axis is called minor axis. 
• The ends points of major axis are called vertices of the ellipse. 
• Length of major axis is 2a 
• length of minor axis is 2b 
• Distance between the foci is 2c 
• Relationship between semi-major axis, semiminor axis and distance of the focus from the centre of the ellipse is a²  = b²  + c²  i.e., c = √(a² - b²)

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 Y²/a² - x²/b² =1.
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 Y²/25 - X²/144 =1.

Let the required equation of hyperbola be \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\)

Length of conjugate axis = 2b 

Given, length of conjugate axis = 3

2b = 3

b = 3/2

b² = 9/4

Distance between foci = 2ae 

Given, distance between foci = 5 

2ae = 5

ae = 5/2

a² e² = 25/4

Now, b² = a² (e² – 1)

b² = a² e² – a²

9/4 = 25/4 - a²

a² = 25/4 - 9/4 

∴ a² = 4

∴ The required equation of hyperbola is \(\frac {x^2}{4} - \frac {y^2}{\frac {9}{4}} = 1\)

i.e., \(\frac {x^2}{4} - \frac {4y^2}{9} = 1\)