**1) The opposite angular points of a square are (3,4)and (1, -1). Then the co-ordinates of other two points are**

(a) **D(1/2 , 9/2) , B(-1/2 , 5/2)** (b) **D(1/2 , 9/2) , B(1/2 , 5/2)**

(c) **D(9/2 , 1/2) , B(-1/2 , 5/2)** (d) **None of these**

(c) Obviously, slope of AC = 5/2 .

Let m be the slope of a line inclined at an angle of 45^{o }to AC,

then tan45^{o} = ±(m-5/2)/(1+m.5/2)

⇒ m = -7/3 , 3/7

Thus, let the slope of AB or DC be 3/7 and that of AD or BC be -7/3

Then equation of AB is 3x-7y=19 = 0.

Also the equation of BC is 7x+3y-4 = 0.

On solving these equations, we get, B(-1/2,5/2)

Now let the coordinates of the vertex D be (h, k).

Since the middle points of AC and BD are same, therefore

1/2(h-1/2) = 1/2(3+1)

⇒ h = 9/2, 1/2(k+5/2) = 1/2(4-1)

⇒ k= 1/2 Hence, D = (9/2 , 1/2)

**2) If the lines y = 3x+1 and 2y x 3 are equally inclined to the line y = mx+4, then m = **

(a) **(1+3√2)/7** (b) **(1-3√2)/7 ** (c) **(1+3√2)/7 ** (d) **(1+5√2)/7**

(d) m_{1} = 3, m_{2} = 1/2 and m_{3} = m

Let the angle between first and third line is θ_{1 }and between second and third is θ_{2} ,then

tan θ_{1 }= (3-m)/(1+3m) and tan θ_{2} = (m-1/2)/(1+m/2)

But θ_{1} = θ_{2} ⇒ (3-m)/(1+3m) = (m-1/2)/(1+m/2)

⇒ 7m^{2}-2m-7

⇒ m = (1±5√2)/7

**3) Let P(-1,0), Q(0,0) and R(3,3√3) be three points. Then the equation of the bisector of the angle PQR is **

(a) **(√3/2)x+y = 0** (b) **x+√3y = 0 ** (c) **√3x+y = 0 ** (d)** x+(√3/2)y = 0**

(c) Slope of QR = (3√3 – 0)/3-0 = √3 i.e., θ = 60

Clearly, ∠ PQR = 120º

OQ is the angle bisector of the angle, so line OQ makes 120º with the positive direction of x-axis.

Therefore, equation of the bisector of

∠PQR is y = tan120º x or y = -√3x

i.e.,√3x+y = 0.

**4) If for a variable line (x/a)+(y/b) = 1, the condition (1/a ^{2})+(1/b^{2})+(1/c^{2}) (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is**

(a) **x ^{2}+y^{2} = c^{2}/2 ** (b)

**x**(c)

^{2}+y^{2}= 2c^{2}**x**(d)

^{2}+y^{2}= c^{2}**x**

^{2}– y^{2}= c^{2}(c) Equation of perpendicular drawn from origin to the line (x/a)+(y/b) = 1 is y-0 = a/b(x-0)

[ m of given line = -b/a ∴ m of perpendicular = a/b]

⇒ by-ax = 0

(x/b)-(y/a) = 1

Now, the locus of foot of perpendicular is the intersection point of line

(x/a)+(y/b) = 1………………………………………….(i)

and

(x/b)-(y/a) = 1……………………………………..(ii)

To find locus, squaring and adding (i) and (ii)

(x/a)+(y/b)^{2} + (x/b)-(y/a)^{2} = 1

⇒ x^{2}(1/a^{2}+1/b^{2}) + y^{2}(1/a^{2}+1/b^{2}) = 1

⇒ x^{2}(1/c^{2}) + y^{2}(1/c^{2}) = 1 [(1/a^{2})+(1/b^{2})+(1/c^{2})]

⇒ x^{2}+y^{2} = c^{2} .

**5) If the circle x ^{2}+y^{2}-4x-6y+k = 0 does not touch or intersect the axes and the point (2, 2) lies inside the circle, then **

(a) **4 < k < 9** (b) **4 < k < 12 ** (c) **9 < k < 12 ** (d) **none of these**

(c) The centre of the given circle is (2, 3) and the radius = √4+9-k

i.e. √13-k .

Since the given circle does not touch or intersect the coordinate axes and the point (2, 2) lies inside the circle

∴ x-cooridnate of centre> radius i.e. 2 > √13-k ,

y-coordinate of centre > radius i.e. 3 > √13-k and

4+4-8-12+k < 0

⇒ 4 > 13-k, 9 > 13-k and -12+k < 0

⇒ k > 9, k > 4 and k < 12 ⇒ 9 < k < 12

**6) If θ _{1} , θ_{2} , be the inclination of tangents drawn from the point P to the circle x^{2}+y^{2} = a^{2} with x-axis, then the locus of P, it is given that cot θ_{1} + cot θ_{2} = c, is **

(a)** c(x ^{2} – a^{2}) = 2xy** (b)

**c(x**(c)

^{2}– a^{2}) = y^{2}– a^{2}**c(y**(d)

^{2}– a^{2}) = 2xy**none of these**

(c) Any tangent is y = mx±a√1+m^{2}

If it passes throughP (h, k) , then (k-mh)^{2} = a^{2}(1+m^{2})

or m^{2}(h^{2}-a^{2})-2mhk+(k^{2}-a^{2}) = 0 , where m = tan θ

cot θ_{1} + cot θ_{2} = c

⇒ (1/m_{1})+(1/m_{2}) = c

⇒ (m_{1}+m_{2})/m_{1}m_{2} = c

⇒ 2hk/(k^{2}-a^{2}) = c

⇒ c(y^{2}-a^{2}) = 2xy

**7) The chord of contact of tangents from a point P to a circle passes through Q. If l _{1} and l_{2} are the lengths of the tangents from P and Q to the circle, then PQ is equal to **

(a) **(I _{1}+I_{2})/2** (b)

**(I**(c)

_{1 }– I_{2})/2**√(I**(d)

_{1}^{2}+I_{2}^{2})**√(I**

_{1}^{2 }– I_{2}^{2})(c) Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two points and x^{2}+y^{2} = a^{2} be the given circle then the chord of contact of tangents drawn from P to the given circle is

xx_{1}+yy_{1} = a^{2}

It will pass through

Q(x_{2}, y_{2}), if x_{2}x_{1}+y_{2}y_{1} = a^{2} …(i)

Now,I_{1} = √(x_{1}^{2}y_{1}^{2}-a^{2}) , I_{2} = √(x_{2}^{2}y_{2}^{2}-a^{2})

⇒ PQ = √(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}

⇒ PQ = √(x_{2}^{2}+y_{2}^{2})+x_{1}^{2}+y_{1}^{2})-2(x_{1}x_{2}+y_{1}y_{2})

⇒ PQ = √(x_{2}^{2}+y_{2}^{2})+x_{1}^{2}+y_{1}^{2})-2a^{2} [using (i)]

⇒ PQ = √(x_{1}^{2}y_{1}^{2}-a^{2})+(x_{2}^{2}y_{2}^{2}-a^{2})

⇒ PQ = √(I_{1}^{2}+I_{2}^{2})

**8) The circles x ^{2}+y^{2}+2g_{1}x-a^{2} = 0 and x^{2}+y^{2}+2g_{2}x-a^{2} = 0 cut each other orthogonally. If p_{1} , p_{2} are perpendicular from (0, a) and (0, –a) on a common tangent of these circle, then p_{1}p_{2} is equal to**

(a)** a ^{2}/2 ** (b)

**a**(c)

^{2}**2a**(d)

^{2}**a**

^{2}+2(b) Given g_{1}g_{2}+a^{2 }…(i)

If lx+my = 1 is common tangent of these circle, then

(-Ig_{1}-1)/√(l^{2}+m^{2}) = ±√(g_{1}^{2}+a^{2})

⇒ (Ig_{1}+1)^{2} = (l^{2}+m^{2})(g_{1}^{2}+a^{2})

⇒ m^{2}g_{1}^{2} – 2Ig_{2}+a^{2}(l^{2}+m^{2}) – 1 = 0

Similarly,m^{2}g_{2}^{2} – 2Ig_{1}+a^{2}(l^{2}+m^{2}) – 1 = 0

So that g_{1} , g_{2 }are the roots of the equation m^{2}g^{2} – 2Ig+a^{2}(l^{2}+m^{2}) – 1 = 0

⇒ g_{1}g_{2} = [a^{2}(l^{2}+m^{2})-1]/m^{2} = -a^{2} (from (i))

a^{2}(l^{2}+m^{2}) = 1 – (a^{2}m^{2}) …(ii)

Now,

p_{1}p_{2} = |ma-1|/√(l^{2}+m^{2}) . |-ma-1|/√(l^{2}+m^{2}) = |1-m^{2}a^{2}|/(l^{2}+m^{2}) = a^{2 }[from (ii)]

**9) The equation of the line touching both the parabolas y ^{2} = 4x and x^{2} = -32y is **

(a) **2x+y-4 = 0 ** (b) **x+2y+4 = 0 ** (c) **x-2y+4 = 0** (d) **x-2y-4 = 0**

(c) Any tangent to y^{2} = 4x is

Solving with x^{2} = -32y

x^{2} = -32[mx+(1/m)]

⇒ x^{2}=32mx=32/m = 0

But the line is tangent to 2nd parabola

Discriminant = 0

⇒ (32)^{2}m^{2} = 4.32/m

⇒ m^{3} = 4.32/(32)^{2} = 1/8 ⇒ m = 1/2

∴ the required common tangent is y = (1/2)x+2

2y = x + 4

x – 2y + 4 = 0.

**10) The algebraic sum of the ordinates of the points of intersection of a parabola and a circle is **

(a) **Proportional to the arithmetic mean of the radius and latus rectum** (b) **zero**

(c) **equal to the ratio of radius and latus rectum** (d) **1**

(b) The equation to a circle is x^{2}+y^{2}+2gx+2fy+c = 0

Any point on the parabola ax y^{2} = 4ax is (at^{2} , 2at) .

If the circle and parabola intersect then

(at^{2})^{2}+(2at)^{2}+2g(at^{2})+2f(2at)+c = 0

a^{2}t^{4}+t^{2}(4a^{2}+2ga)+4fat+c = 0

This is a 4th degree equation in t giving 4 roots t_{1}, t_{2}, t_{3}, t_{4} .

∴ circle and parabola meet in 4 points

t_{1}+t_{2}+t_{3}+t_{4} = coefficient of t_{3} / coefficient of t_{4} = 0

2at_{1}+2at_{2}+2at_{3}+2at_{4} = 0

Algebraic sum of the ordinates of the points of intersection = 0.

**11) The directrix of the parabola 4x = 2y ^{2} – 2y has the equation **

(a) **4x – 1 = 0** (b) **4x+5 = 0 ** (c) **4x – 5 = 0 ** (d) **4y+5 = 0**

(b) y^{2}-2y = 4x

(y-1)^{2}-1 = 4x

(y-1)^{2} = 4x+1 = 4[x+(1/4)]

Parabola with vertex (-1/4 , 1)

4a = 4 ⇒ a = 1

x+(1/4) = -1

x = -5/4

Equation to the directrix is x = -5/4

4x+5 = 0

**12) Tangents are drawn from the point P(–2, 6) to the parabola y ^{2} = 8x meet it at A and B. Then the locus of the centre of the circle described on AB as diameter is **

(a) **y ^{2} = x – 2** (b)

**y**(c)

^{2}= 2(x-2)**y**(d)

^{2}= 3(x-2)**y**

^{2}= 4(x -2)(d) 4a = 8 ⇒ a = 8

Directrix is x = – a ⇒ x = – 2

∴ P is a point on the directrix.

Theory: Perpendicular tangents intersect on the directrix.

∴ tangents at A (t_{1}) and B (t_{2}) are perpendicular

∴ (1/t_{1})×(1/t_{2}) = -1

Now mid-point of AB = [2(t_{1}^{2}+t_{2}^{2})/2 , 4(t_{1}+t_{2})/2] = [t_{1}^{2}+t_{2}^{2} , 2(t_{1}+t_{2})]

This is the required centre.

Let it be (α, β)

α = t_{1}^{2}+t_{2}^{2 }; β = 2(t_{1}+t_{2})

Now β^{2} = 4(t_{1}^{2}+t_{2}^{2}+2t_{1}t_{2})

Locus of (α, β) is y^{2} = 4(x-2)

**13) Let P be a variable point on (x ^{2}/16)+(y^{2}/9) = 1, then the maximum area of SPS’ , where S and S’ are the foci is**

(a) **3√7 ** (b) **2√7 ** (c) **√7 ** (d) **4√7**

(a) a^{2} = 16 ; b^{2} = 9

b^{2} = a^{2}(1-e^{2}) ⇒ 1-e^{2} = b^{2}/a^{2} = 9/16

⇒ e^{2} = 1-(9/16) = 7/16

⇒ e = √7/4

Area of ΔSPS’ = 1/2 SS’ . PM = 1/2 2ae×3sinθ

= (1/2)2×4(√7/4 ) = 3√7 sinθ

Maximum value of sin θ = 1

Maximum area = 3√7

**14) The locus of the point of intersection of the tangents at the ends whose eccentric angle differ by α is **

(a) **(x ^{2}/a^{2})+(y^{2}/b^{2}) = tan^{2} α ** (b)

**(x**

^{2}/a^{2})+(y^{2}/b^{2}) = sec^{2}α(c) **(x ^{2}/a^{2})+(y^{2}/b^{2}) = cos^{2} α** (d)

**(x**

^{2}/a^{2})+(y^{2}/b^{2}) = sin^{2}α(b) Let θ and Φ be the eccentric angels of the two points.

Given Φ – θ = α

Φ = α+θ

Equation to the tangent at θ

x/a cosθ+y/b sinθ = 1 … (i)

II tangent is x/a cos(α+θ)+y/b sin(α+θ) = 1 … (ii)

cosα(x/a cosθ+y/b sinθ)+sinα(-x/a sinθ+y/b cosθ) = 1

i.e. cosα+sinα(-x/a sinθ+y/b cosθ) =

∴ -x/a sinθ+y/b cosθ = (1-cosα)/sinα = tanα/2

Already x/a cosθ+y/b sinθ = 1

Squaring and adding, (x^{2}/a^{2})+(y^{2}/b^{2}) = sec^{2} α/2

**15) A tangent to the ellipse (x ^{2}/a^{2})+(y^{2}/b^{2}) = 1 has slope 2 and makes an intercept √5 units on the auxiliary circle, then a focus of the ellipse is **

(a) **(5/2 , 0) ** (b) **(4,0)** (c) **(2,0)** (d) **(5,0)**

(a) Equation of tangent is y = 2x+√(a^{2}4+b^{2})

⇒ y = 2x+√(4a^{2}+b^{2})

Auxiliary circle is x^{2}+y^{2} = a^{2} ; centre (0, 0) ; radius a

Let M be the mid-point AM^{2} = a^{2}-[√(4a^{2}+b^{2})/√(1+2^{2})]^{2}

AM^{2} = a^{2}-(4a^{2}+b^{2})/5

= (5a^{2}-4a^{2}-b^{2})/5 = (a^{2}-b^{2})/5

[a^{2}-a^{2}(1-e^{2})]/5 = a^{2}e^{2}/5

ae = 5/2 ⇒ focus (5/2 , 0)

**16) The sum of the reciprocals of the slopes of the tangents drawn from any point to the ellipse (x ^{2}/a^{2})+(y^{2}/b^{2}) = 1 is 3, then the locus of the point is **

(a) **3y ^{2} – 2xy – 3b^{2} = 0 ** (b)

**3y**(c)

^{2}+2xy – 3b^{2}= 0**y**(d)

^{2}+xy – 3b^{2}= 0**2y**

^{2}+xy – b^{2}= 0(a) Any tangent to the ellipse is y = mx+√(a^{2}m^{2}+b^{2})

Let it pass through (α, β)

(β – mα)^{2} = a^{2}m^{2}+b^{2}

⇒ m^{2}(α^{2} – a^{2})-2mαβ+β^{2} – b^{2} = 0

m_{1}m_{2} are the slopes

m_{1}+m_{2} = 2αβ/(α^{2} – a^{2})

m_{1}m_{2} = (β^{2} – b^{2})/(α^{2} – a^{2})

(1/m_{1})+(1/m_{2}) = 3 ⇒ (m_{1}+m_{2})/m_{1}m_{2} = 3

⇒ [2αβ/(α^{2} – a^{2})]/[(β^{2} – b^{2})/(α^{2} – a^{2})]

⇒ 2αβ/(β^{2} – b^{2}) ⇒ 3β^{2} – 3b^{2} = 2αβ

Locus of (α, β) is 0 3y^{2} – 2xy – 3b^{2} = 0

**17) The asymptotes of xy = 4x+3y are **

(a) **x = 4 y = 4** (b) **x = 4 y = 3** (c) **x = 3 y = 4** (d) **x = 3 y = 3 **

(c) We know that the combined equation of the asymptotes and the equation of the hyperbola differ by a constant.

∴ we can take the equations of the asymptotes as xy-4x-3y+k = 0

We have to find k such that the above equation a pair of straight lines

∴ abc+2fgh – af^{2} – bg^{2} – ch^{2} = 0

i.e., 2(-3/2)(-2)(1/2) – k(1/2)^{2} = 0

⇒ k = 12

Combined equation of the asymptotes

xy-4x-3y+12 = 0

x(y-4)-3(y-4) = 0

⇒ (x-3)(y-4) = 0

**18) If PN is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, the mid-point of PN is**

(a) **circle** (b) **parabola** (c) **ellipse** (d) **hyperbola**

(d) Let xy = c^{2} be the rectangular hyperbola, and let P (x_{1}, y_{1}) be a point on it.

Let Q(h, k) be the mid-point of PN. Then the coordinates of Q are (x_{1} , y_{1}/2) .

∴ x_{1} = h and y_{1}/2 = k

⇒ x_{1} = h and y_{1} = 2k

But (x_{1}, y_{1}) lies on xy = c^{2}.

∴ h (2k) = c^{2} ⇒ hk = c2/2

Hence, the locus of (h, k) is xy = c^{2}/2, which is hyperbola

**19) The locus of the mid-points of the chords of the circle x ^{2}+y^{2} = 16 which are tangent to the hyperbola 9x^{2} – 6y^{2} = 144 is**

(a) **(x ^{2}+y^{2})^{2} = 16x^{2} – 9y^{2} ** (b)

**(x**

^{2}+y^{2})^{2}= 9x^{2}– 16y^{2}(c) **(x ^{2 }– y^{2})^{2} = 16x^{2} – 9y^{2} ** (d)

**none of these**

(a) The given hyperbola is (x^{2}/16) – (y^{2}/9) …(i)

Any tangent to (i) is y = mx+√(16m^{2} – 9) …(ii)

Let (x_{1},y_{1}) be the mid-point of the chord of the circle x^{2}+y^{2} = 16

Then equation of the chord is T = S_{1}

i.e. xx_{1}+yy_{1} – (x_{1}^{2}+y_{1}^{2}) = 0 …(iii)

Since (ii) and (iii) represent the same line.

∴ m/x_{1} = -1/y_{1} = √(16m^{2}-9)/-(x_{1}^{2}+y_{1}^{2})

⇒ m = -(x_{1}/y_{1}) and (x_{1}^{2}+y_{1}^{2})^{2} = y_{1}^{2}(16m^{2}-9)

⇒ (x_{1}^{2}+y_{1}^{2})^{2} = 16×(x_{1}^{2}/y_{1}^{2})y_{1}^{2} – 9y_{1}^{2} = 16x_{1}^{2} – 9y_{1}^{2}

∴ locus of (x_{1},y_{1}) is (x^{2}+y^{2})^{2} = 16x^{2} – 9y^{2}

**20) The slope of the common tangents to the hyperbola (x ^{2}/9^{2}) – (y^{2}/16^{2}) = 1 and (y^{2}/9^{2}) – (x^{2}/16^{2}) = 1**

(a) **-2,2** (b) **-1,1** (c) **1,2 ** (d) **2,1**

(b) Given hyperbolas are (x^{2}/9)-(y^{2}/16) = 1 …(i)

and (y^{2}/16)-(x^{2}/16) = 1 …(ii)

Any tangent to (i) y = mx±√(9m^{2}-16) …(iii)

Putting in (ii), we get 16[mx±√(9m^{2}-16)]^{2} – 9x^{2}

⇒ (16m^{2} – 9)x^{2} ± 32m(√(9m^{2}-16))x+144m^{2}-256-144 = 0

⇒ (16m^{2} – 9)x^{2} ± 32m(√(9m^{2}-16))x+(144m^{2} – 400) …(iv)

If (iii) is a tangent to (ii), then the roots of (iv) are real and equal.

∴ Disc. = 0

32×32m^{2}(9m^{2}-16) = 4(16m^{2} – 9)(144m^{2} – 400) = 64(16m^{2} – 9)(9m^{2}-25)

⇒ 16m^{2}(9m^{2}-16) = (16m^{2} – 9)(9m^{2}-25)

⇒ 144m^{4} – 256m^{2} = 144m^{4} – 481m^{2}+225

⇒ 225m^{2} = 225

⇒ m^{2} = 1 ⇒ m = ±1