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Correct option (b)  1/2

Explanation :

Let M = (0, k) so that the slope of LM is

2 - k/t - 0 = -t

⇒ 2 - k = -t²   ....(1)

Let (x, y) be the midpoint of LM. Therefore

x = t/2 and y = 2 + k/2

Hence, from Eq. (1), we have

2y = 2 + k = 2 + (2 + t²)

= 4 + t²

⁼ 4 + 4x²

y = 2 + 2x²

x² = 1/2(y - 2)

Hence, the latus rectum is 1/2.

Let P(h, k) be a point in the plane of y² = 4ax. It is known that the normal to parabola at (at², 2at) is tx + y = 2at + at³. This normal passes through P(h,k

th + k = 2at + 2at + at³ 

at³ + (2a - h)t - k = 0   .....(1)

Equation (1) is a cubic equation in t and hence, in general, it has three roots and hence, there are three points on the curve at which normals drawn to the parabola are concurrent at P(h, k). 

Let P(x₁, y₁) be a point on the line 2x - 3y + 4 = 0. Therefore

2x₁ - 3y₁ + 4 = 0 ....(1)

Now, the chord of contact of (x₁, y₁) with respect to y² + 4ax is

yy₁ - 2a(x + x₁) = 0 ....(2)

From Eqs. (1) and (2), we get

yy₁ - 2ax - a(3y₁ - 4) = 0

y₁(y - 3a) 2a(x - 2) = 0  ...(3)

Eq. (3) represents the lines passing through the fixed point which is the intersection of the lines x = 2 and y = 3a. Hence, the fixed point is (2, 3a).

Suppose (x₁, y₁) is the point whose chord of contact with respect to y²  = 4x is

2x - 7y + 2 = 0  ...(1)

However, the chord of contact of (x₁, y₁) is

S ≡ yy₁ - 2(x + x₁) = 0

Therefore, from Eqs. (1) and (2), we get

-2/2 = y₁/-7 = -2x₁/2

Hence, x₁ = 1 and y₁ = 7. Thus, the required point is (1, 7).

Correct option  (D)  90°

Explanation 

 Let P be (x₁, y₁) so that

x₁ + 4a = 0 .....(1)

Chord of contact of P(x₁,y₁) is

yy₁ - 2a(x + x₁) = 0

⇒ yy₁ - 2ax + 8a² = 0

[∴ x₁ = -4a  from Eq.(1)]

⇒ yy₁ - 2a(x + x₁) = 0

⇒ yy₁ -2ax + 8a² = 0

⇒ 2ax - y₁y_/8a² = 1

Hence, the combined equation of the pair of lines OQ and OR is

y² - 4ax(2ax - y₁y/8a²)

In this equation, the coefficient of x² + the coefficient of y² = −1 + 1 = 0. Hence ∠QOR  = 90°

Correct option (A)  y² = 4ax + 3(x+ a)²

Explanation :

According to Problem 28 of the section ‘Subjective Problems’, the locus of the point of intersection of the tangents to the parabola y² = 4ax which include angle α is

cot² α(y² - 4ax) = (x +  a)²

Now, substituting  α = 60° the required locus is

1/3(y - 4ax) = (x + a)²

y² - 4ax = 3(x + a)²

Concept:

The standard equation of an ellipse:

\(\rm {x^2\over a^2}+{y^2\over b^2} = 1\)

Where 2a and 2b are the length of the major axis and minor axis respectively and center (0, 0)

The eccentricity = \(\rm \sqrt{(a^2-b^2)}\over a\)

Length of latus recta = \(\rm 2b^2 \over a\)

Distance from center to focus = \(\rm \sqrt{a^2-b^2}\)

Calculation:

Given ellipse \(\rm {x^2\over100}+{y^2\over64} = 1\)

The eccentricity (e)

⇒ e = \(\rm \sqrt{100-64}\over 10\)

⇒ e = \(\rm \sqrt{36}\over 10\)

⇒ e = 0.6

Now distance between foci = 2ae

= 2 × 10 × 0.6

∴ Distance between foci = 12

Correct option  (i)(c)(ii)(b)(iii)(d)

Explanation :

(i) The parabola is

Y² = 4(1/2)(x)

where X = x - 1 and Y = y - 1. Therefore, the vertex is

(X = 0, Y = 0) = (x - 1 = 0, y -  1 =  0) = (1, 1)

 (ii)  y² = 4(x -1) is Y² = 4X where X = x - 1 and Y = y and a - 1. The focus is

(X = 1, Y = 0) =(x - 1 = 1,y = 0) = (2,0)

 (iii)  (y - 2)² = 4(x - 1)  ⇒ Y² = 4X where X = x - 1 and Y = y − 2. The directrix equation is

X = -a = -1

⇒ x - 1 = -1

⇒ x = 0

Tangent at (at², 2at) is ty = x + at² . Thus the slope of the tangent at t is 1/t. Hence, the equation of the normal at (at², 2at) is

 y - 2at = -t(x - at²)

⇒ tx + y = 2at + at³

The given equation is written as

(x - 2)² = 1 - 2y = -2(y - 1/2)

Let X = x - 2 and Y = y (1/2). Substituting in above equation we get

X² = -4aY

where a = 1/2.

The directrix equation Y = a or y − 1 = 0. Therefore, the director circle is the directrix y − 1 = 0.

n: The equation of the curve is

y² = (8/3)x = 4(2/3)x

so that a = 2/3. Hence, S = (a,0) = (2/3,0) is the focus, 

x + 2/3 = 0 or 3x + 2 = 0

is the equation of the directrix and 4/83 a  is the length of the latus rectum.

Correct option (C) L = 0 touches the parabola S = 0

Explanation :

Substituting y = 1 x in the equation of the parabola, we get

1 - x - x + x² = 0

⇒ x² - 2x + 1 = 0

⇒ (x - 1)² = 0

Hence, L = 0 touches the parabola at (1, 1).

Correct option  (D) ln = am² 

Explanation :

Suppose lx + my + n = 0 touches the parabola y² = 4ax at (at², 2at). However, ty = x + at² is the equation of the tangent at (at², 2at). Therefore

l/1 = m/-t = n/at²

⇒ - m/l = t = -n/ma

⇒ am² = ln

Correct option (B)  24

Explanation :

Suppose 2x + y + k = 0 is normal at (-2t², -4t). However, the normal at (-2t², -4t) is

tx + y = -4t - 2t³

⇒ tx + y + 4t + 2t³ = 0

Therefore

t/2 = 1/1 = 4t + 2t³/k

⇒ t = 2 and k = 24

Let the tangents at t₁ and t₂ make complementary angles, say α and β, with the axis of the parabola. Therefore

tan α = 1/t₁

and  tan β = 1/t₂

and suppose α + β = 90º ⇒ t₁t₂ = 1. Since the abscissa of the point of intersection of the tangent at t₁ and t₂ is at₁t₂ = a, the point of intersection lies on the line x = a.

Correct option   (D)  b = −1, c = 1

Explanation :

The point (1, 1) lies on the parabola, which implies that

b + c = 0  ...(1)

 Also, y = x touches the parabola

⇒ the quadratic x² + (b - 1) x + c = 0 has equal roots

⇒ (b - 1)² - 4c = 0

Therefore, from Eq. (4.108), we get

(b - 1)² + 4b = 0

⇒(b + 1)² = 0

⇒b = -1 and c = 1

Concept:

Ellipse:

Calculation:

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

Compare with the standard equation \(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

So, a² = 16 and b² = 25 ⇔ a = 4 and b = 5 (b > a)

So, eccentricity = \(\rm \sqrt{1-\frac{a^2}{b^2}}\)

= \(\sqrt{1-\frac{16}{25}}\)

= 3/5

Concept:

Equation of ellipse: \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Foci: (±ae, 0)

Eccentricity, e = \(\rm \sqrt{1-\frac{b^2}{a^2}}\)

Calculation:

Here,foci = (±2,0) = (±ae, 0) and the eccentricity, e = 1/2

ae = 2 

⇒ a × 1/2 = 2 

⇒ a = 4

⇒ a² = 16

Now, e = \(\rm \sqrt{1-\frac{b^2}{a^2}}\)

⇒ b²= a²(1 - e²)

= 16(1 - 1/4)

= 16× (3/4)

= 12

∴ Equation of ellipse = \(\rm \frac{x^2}{16}+\frac{y^2}{12}=1\)

Hence, option (3) is correct.

CONCEPT:

The properties of a rectangular hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) are:

• Its centre is given by: (0, 0)
• Its foci are given by: (- ae, 0) and (ae, 0)
• Its vertices are given by: (- a, 0)  and (a, 0)
• Its eccentricity is given by: \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)
• Length of transverse axis = 2a and its equation is y = 0.
• Length of conjugate axis = 2b and its equation is x = 0.
• Length of its latus rectum is given by: \(\frac{2b^2}{a}\)

CALCULATION:

Here, we have to find the equation of the hyperbola whose foci are (± 5, 0) and the conjugate axis is of length 8.

By comparing the foci (± 5, 0) with (± ae, 0)

⇒ ae = 5

∵ Length of the conjugate axis is given by 2b

⇒ 2b = 8

⇒ b = 4

As we know that, eccentricity of a hyperbola is given by \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)

⇒ a²e² = a² + b²

⇒ 25 = a² + 16

⇒ a² = 9

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

Hence, option A is the correct answer.

Concept:

The equation of the hyperbola is \(\rm \dfrac {y^2}{b^2}- \dfrac{x^2}{a^2} = 1\) with the foci (0 , ± ae) 

Length of the transverse axis =  2a

Calculations: 

Since the coordinates of the one focus at (0, 4) = (0 , ± ae) , it is a case of vertical hyperbola

⇒ ae = 4

It is a case of vertical hyperbola

⇒ The equation of hyperbola is \(\rm \dfrac {y^2}{b^2}- \dfrac{x^2}{a^2} = 1\) ....(1) 

 Length of the transverse axis = 6

⇒ 2a=6

⇒  a = 3

\(\rm \text {Also}\;\;a^2e^2 = a ^ 2+ b^2\)

⇒\(\rm 4^2 =3 ^ 2+ b^2\)

⇒\(\rm b^2 = 7\)

Equation (1) becomes 

\(\rm \frac {y^2} 7 - \frac {x^2} 9 = 1\)

Hence, The equation of the hyperbola with center at the origin, length  of the transverse axis is 6 and one focus at (0, 4) is \(\rm \frac {y^2} 7 - \frac {x^2} 9 = 1\) 

Concept:

Standard equation of an hyperbola : \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) 

• Coordinates of foci = (± ae, 0)
• Eccentricity (e) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) ⇔ a2e2 = a2 + b2
• Length of Latus rectum = \(\rm \frac{2b^2}{a}\)

Calculation:

Given: \(\rm \frac{x^2}{100} - \frac{y^2}{75} = 1\)

Compare with the standard equation of a hyperbola: \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} - \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\)

So, a2 = 100 and b2 = 75

Now, Eccentricity (e) = \(\sqrt {1 + {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) 

= \(\sqrt {1 + \frac{75}{100}}\)

= \(\sqrt {1 + \frac{3}{4}}\)

= \(\sqrt { \frac{7}{4}}\)

Concept:

The standard equation of a hyperbola:

\(\rm {(x-h)^2\over a^2}-{(y-k)^2\over b^2} = 1\)

where 2a and 2b are the length of the transverse axis and conjugate axis respectively and centre (h, k)

Note: The centre is the midpoint of the 2 foci.

The eccentricity = \(\rm \sqrt{a^2+b^2}\over a\)

Length of latus recta = \(\rm 2b^2 \over a\)

Distance from center to focus = \(\rm \sqrt{a^2+b^2}\)

Calculation:

Given foci are (5,0) and (-3,0)

Center = \(\rm \left({5+(-3)\over2},{0+0\over2}\right)\) = (1, 0)

Now distance of focus from the center = \(\rm \sqrt{a^2+b^2}\)

⇒ \(\rm \sqrt{(5-1)^2+(0-0)^2} = \rm \sqrt{a^2+b^2}\)

⇒ a² + b² = 16               ...(i)

The eccentricity = \(\rm \sqrt{a^2+b^2}\over a\)

⇒ 2 = \(\rm \sqrt{16}\over a\) 

⇒ a = 2

Putting it in (i)

⇒ 4 + b² = 16

⇒ b² = 12

a² = 4

The equation of the hyperbola

\(\rm {(x-h)^2\over a^2}-{(y-k)^2\over b^2} = 1\)

⇒ \(\rm {(x-1)^2\over 4}-{(y-0)^2\over 12} = 1\)

⇒ \(\boldsymbol{\rm {(x-1)^2\over 4}-{y^2\over 12} = 1}\)

Concept:

The distance between the centre and the focus of an ellipse is c = ae

The equation of an ellipse with the length of the major axis 2a and the minor axis 2b is given by: \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).

Calculation:

Length of the minor axis = 2b = 8.

⇒ b = 4.

Also, c = distance between the centre and the focus = ae = 2.

c2 = a2e2 = a2 - b2

∴ 22 = a2 - 42

⇒ a² = 20

Equation of the ellipse = \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).

⇒ \(\rm \frac{x^2}{20}+\frac{y^2}{16}=1\).

Let t₁, t₂ and t₃ be the parameters of the feet of the normals drawn from (h, 0). Hence, t₁, t₂ and t₃ are the roots of the cubic equation t³ + (2 - h)t = 0. That is, t₁, t₂ and t₃ are the roots of t(t² + 2 - h) = 0 Since the equation has three roots, h must be greater than 2. 

Concept:

The distance between a focus of the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and its center is denoted by c and is equal to c² = a² - b².

The distance between a focus of a hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its center is denoted by c and is equal to c2 = a2 + b², and its eccentricity e is equal to \(\rm \frac{c}{a}\).

Calculation:

The distance between the focus and the center of the ellipse \(\rm \frac{x^2}{25}+\frac{y^2}{16}=1\) is c² = 25 - 16 = 9 

⇒ c = 3.

Since the hyperbola has the same foci, its c will be the same as that of the ellipse.

Using e = \(\rm \frac{c}{a}\) for the hyperbola, we get:

3 = \(\rm \frac{3}{a}\)

⇒ a = 1

Using c2 = a2 + b², we get:

32 = 12 + b2

⇒ b² = 8.

The equation of the hyperbola is \(\rm \frac{x^2}{1}-\frac{y^2}{8}=1\).

Concept:

The eccentricity of the curve \(\rm \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) is \(\rm e^2 = 1-\left({b^2\over a^2}\right)\)

The eccentricity of the curve \(\rm \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\) is \(\rm e^2 = 1+\left({b^2\over a^2}\right)\)

Foci of Hyperbola and Ellipse are (ae, 0) and (4e, 0)

Calculation:

For given Hyperbola \(\rm \frac {x^2}{144} - \frac {y^2}{81} = \frac {1}{25}\)

⇒ \(\rm \frac {25}{144}x^2 - \frac {25}{81}y^2 = 1\)

∴ \(\rm {a_h}^2 = \frac{144}{25}\) and \(\rm {b_h}^2 = \frac{81}{25}\)

Eccentricity of hyperbola 

\(\rm {e_h}^2 = 1+\left({{b_h}^2\over {a_h}^2}\right)\)

⇒ \(\rm {e_h}^2 = 1+\left({81\over 144}\right)\)

⇒ \(\rm {e_h}^2 = \frac{225}{144}⇒\boldsymbol{\rm e_h=\frac{15}{12}}\)

Focus of hyperbola Fh = (aheh, 0), Where eh is the eccentricity of the hyperbola.

⇒ Fh = \(\rm \left(\left(\frac{12}{5}\times\frac{15}{12}\right), 0\right)\) 

⇒ Fh = (3, 0)

For given Ellipse \(\rm\frac {x^2}{16} + \frac {y^2}{b^2} =1\)

∴ \(\rm {a_e}^2 = 16\) and \(\rm {b_e}^2 = b^2\)

Focus of ellipse F_e = (ae, 0) = (4e_e, 0), Where e_e is eccentricity of the ellipse.

Given Focus of ellipse F_e = F_h 

⇒ (4e_e, 0) = (3, 0)

⇒ 4e_e = 3 ⇒ e_e = \(\rm \boldsymbol {3\over4}\)

Also Eccentricity of an ellipse

\(\rm {e_e}^2 = 1-\left({{b_e}^2\over {a_e}^2}\right)\)

⇒ \(\rm \left({3\over4}\right)^2 = 1-\left({{b}^2\over {4}^2}\right)\)

⇒ \(\rm 1-{9\over16} = {{b}^2\over16}\)

⇒ b² = 7

Correct option  (B) x + 2 = 0

Explanation :

Orthocentre of a triangle formed by three tangents to the parabola y² = 4ax lies on the directrix x + a = 0 (see Problem 4 in the section ‘Subjective Problems’). Here, a = 2 gives that the orthocentre lies on the line x = 2 = 0.

The line

y = mx + a/m and y = -x/m - am

are tangents to the parabola and they intersect at right angles. Their point of intersection satisfies the equation 

x(m + 1/m) = -a(m + 1/m)

Hence, the locus of the point of intersection is the line x −a which is the directrix.

Let the tangents be

t_ry = x + at_r²

where r = 1, 2 and 3. By Problem 1, the feet of the perpendiculars drawn from the focus of the parabola onto the three tangents (which are the sides of a triangle) are collinear on the tangent at the vertex. Hence, from the section ‘Pedal Line (or Simson’s Line)’ , the circumcircle of the triangle passes through the focus.

Correct option (C), (D)

Explanation :

If the normals at (at₁², 2at₁) and (at₂², 2at₂) meet again on the curve, then, x₁x₂ = 4a² and  y₁y₂ = 8a². Here, a = 2.

Calculation:

So, its parametric coordinate is (at², 2at)

We can check it by putting the above point in the equation of parabole 

L.H.S = y2

⇒ (2at)²

⇒ 4a²t²     -----(i)

R.H.S = 4ax

⇒ 4 × a × at²

⇒ 4a2t2     -----(ii)

From (i) and (ii), we get 

∴ (at², 2at) is the parametric coordinates of the parabola y2  = 4ax.

Correct option (D)  90º

Explanation :

Let P be (at², 2at) so that t = 2 because it lies on the line y = x. Suppose normal at P meets the curve at Q = (at'² , 2at'). So

t'² = -t - 2/t = -2 - 1 (∴ t -= 2)

Therefore, Q = (9a, - 6a) and P = (4a, 4a). Now

Slop of SP x Slop of SQ = (4a/4a - a)(6a/a - 9a)

= 4/3(-6/8) = -1

Hence,∠PSQ = 90º

The equation of the chord in terms of its midpoint M(x₁, y₁) is

S₁ = S₁₁

⇒ yy₁ - 2a(x + x₁) = y₁² - 4ax₁

yy₁ - 2ax = y₁² - 2ax₁ which passes through (0,0)

y₁² - 2ax₁ = 0

Hence, the locus of M(x₁, y₁) is the parabola y² = 2ax.

We have the equations

x = t² +1 ...(1)

y = 2t + 1 ...(2)

From (1) and (2), we get

(y – 1)² = 4 (x – 1) ...(3)

Let Y = y – 1 and  X = x – 1

Then from equation (3) we have

⇒ Y² = 4X

Equation of the directrix for parabola Y² = 4X is:

X = –1

⇒ x – 1 = –1

⇒ x = 0

This is the required equation of directrix.

Concept:

The equation of the hyperbola is:

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

The vertices are (± a, 0)

The asymptotes are the straight lines:

y = (b/a)x and y = -(b/a)x.

Given:

Vertices are (± 5, 0)       

Asymptotes are: 3x ± 5y = 0

Analysis:

5y = 3x & 5y = -3x

\(y = \left( {\frac{3}{5}} \right)x\;\& \;y = - \left( {\frac{3}{5}} \right)x\) 

∴ a = 5 and b = 3

Equation of hyperbola:

\(\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{9} = 1\) 

9x² – 25y² = 225

The given equation is written as

(y - 1)² = -4(x + 1/2)

Let X = x + 1/2 and Y = y -1

Then we have

Y² = -4X 

 Therefore, the vertex is given by

(X = 0,Y = 0) = (-1/2, 1)

 Since a = 1, the focus (X = −1, Y = 0) is

X = a or x + 1/2 = 1 or x - 1/2 = 0

 The latus rectum is given by 4a = 4