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 45o to AC,
then tan45o = ±(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) m1 = 3, m2 = 1/2 and m3 = m
Let the angle between first and third line is θ1 and between second and third is θ2 ,then
tan θ1 = (3-m)/(1+3m) and tan θ2 = (m-1/2)/(1+m/2)
But θ1 = θ2 ⇒ (3-m)/(1+3m) = (m-1/2)/(1+m/2)
⇒ 7m2-2m-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/a2)+(1/b2)+(1/c2) (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is
(a) x2+y2 = c2/2 (b) x2+y2 = 2c2 (c) x2+y2 = c2 (d) x2 – y2 = c2
(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
⇒ x2(1/a2+1/b2) + y2(1/a2+1/b2) = 1
⇒ x2(1/c2) + y2(1/c2) = 1 [(1/a2)+(1/b2)+(1/c2)]
⇒ x2+y2 = c2 .
5) If the circle x2+y2-4x-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 θ1 , θ2 , be the inclination of tangents drawn from the point P to the circle x2+y2 = a2 with x-axis, then the locus of P, it is given that cot θ1 + cot θ2 = c, is
(a) c(x2 – a2) = 2xy (b) c(x2 – a2) = y2 – a2 (c) c(y2 – a2) = 2xy (d) none of these
(c) Any tangent is y = mx±a√1+m2
If it passes throughP (h, k) , then (k-mh)2 = a2(1+m2)
or m2(h2-a2)-2mhk+(k2-a2) = 0 , where m = tan θ
cot θ1 + cot θ2 = c
⇒ (1/m1)+(1/m2) = c
⇒ (m1+m2)/m1m2 = c
⇒ 2hk/(k2-a2) = c
⇒ c(y2-a2) = 2xy
7) The chord of contact of tangents from a point P to a circle passes through Q. If l1 and l2 are the lengths of the tangents from P and Q to the circle, then PQ is equal to
(a) (I1+I2)/2 (b) (I1 – I2)/2 (c) √(I12+I22) (d) √(I12 – I22)
(c) Let P(x1, y1) and Q(x2, y2) be two points and x2+y2 = a2 be the given circle then the chord of contact of tangents drawn from P to the given circle is
xx1+yy1 = a2
It will pass through
Q(x2, y2), if x2x1+y2y1 = a2 …(i)
Now,I1 = √(x12y12-a2) , I2 = √(x22y22-a2)
⇒ PQ = √(x2-x1)2+(y2-y1)2
⇒ PQ = √(x22+y22)+x12+y12)-2(x1x2+y1y2)
⇒ PQ = √(x22+y22)+x12+y12)-2a2 [using (i)]
⇒ PQ = √(x12y12-a2)+(x22y22-a2)
⇒ PQ = √(I12+I22)
8) The circles x2+y2+2g1x-a2 = 0 and x2+y2+2g2x-a2 = 0 cut each other orthogonally. If p1 , p2 are perpendicular from (0, a) and (0, –a) on a common tangent of these circle, then p1p2 is equal to
(a) a2/2 (b) a2 (c) 2a2 (d) a2+2
(b) Given g1g2+a2 …(i)
If lx+my = 1 is common tangent of these circle, then
(-Ig1-1)/√(l2+m2) = ±√(g12+a2)
⇒ (Ig1+1)2 = (l2+m2)(g12+a2)
⇒ m2g12 – 2Ig2+a2(l2+m2) – 1 = 0
Similarly,m2g22 – 2Ig1+a2(l2+m2) – 1 = 0
So that g1 , g2 are the roots of the equation m2g2 – 2Ig+a2(l2+m2) – 1 = 0
⇒ g1g2 = [a2(l2+m2)-1]/m2 = -a2 (from (i))
a2(l2+m2) = 1 – (a2m2) …(ii)
Now,
p1p2 = |ma-1|/√(l2+m2) . |-ma-1|/√(l2+m2) = |1-m2a2|/(l2+m2) = a2 [from (ii)]
9) The equation of the line touching both the parabolas y2 = 4x and x2 = -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 y2 = 4x is
Solving with x2 = -32y
x2 = -32[mx+(1/m)]
⇒ x2=32mx=32/m = 0
But the line is tangent to 2nd parabola
Discriminant = 0
⇒ (32)2m2 = 4.32/m
⇒ m3 = 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 x2+y2+2gx+2fy+c = 0
Any point on the parabola ax y2 = 4ax is (at2 , 2at) .
If the circle and parabola intersect then
(at2)2+(2at)2+2g(at2)+2f(2at)+c = 0
a2t4+t2(4a2+2ga)+4fat+c = 0
This is a 4th degree equation in t giving 4 roots t1, t2, t3, t4 .
∴ circle and parabola meet in 4 points
t1+t2+t3+t4 = coefficient of t3 / coefficient of t4 = 0
2at1+2at2+2at3+2at4 = 0
Algebraic sum of the ordinates of the points of intersection = 0.
11) The directrix of the parabola 4x = 2y2 – 2y has the equation
(a) 4x – 1 = 0 (b) 4x+5 = 0 (c) 4x – 5 = 0 (d) 4y+5 = 0
(b) y2-2y = 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 y2 = 8x meet it at A and B. Then the locus of the centre of the circle described on AB as diameter is
(a) y2 = x – 2 (b) y2 = 2(x-2) (c) y2 = 3(x-2) (d) y2 = 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 (t1) and B (t2) are perpendicular
∴ (1/t1)×(1/t2) = -1
Now mid-point of AB = [2(t12+t22)/2 , 4(t1+t2)/2] = [t12+t22 , 2(t1+t2)]
This is the required centre.
Let it be (α, β)
α = t12+t22 ; β = 2(t1+t2)
Now β2 = 4(t12+t22+2t1t2)
Locus of (α, β) is y2 = 4(x-2)
13) Let P be a variable point on (x2/16)+(y2/9) = 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) a2 = 16 ; b2 = 9
b2 = a2(1-e2) ⇒ 1-e2 = b2/a2 = 9/16
⇒ e2 = 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) (x2/a2)+(y2/b2) = tan2 α (b) (x2/a2)+(y2/b2) = sec2 α
(c) (x2/a2)+(y2/b2) = cos2 α (d) (x2/a2)+(y2/b2) = sin2 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, (x2/a2)+(y2/b2) = sec2 α/2
15) A tangent to the ellipse (x2/a2)+(y2/b2) = 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+√(a24+b2)
⇒ y = 2x+√(4a2+b2)
Auxiliary circle is x2+y2 = a2 ; centre (0, 0) ; radius a
Let M be the mid-point AM2 = a2-[√(4a2+b2)/√(1+22)]2
AM2 = a2-(4a2+b2)/5
= (5a2-4a2-b2)/5 = (a2-b2)/5
[a2-a2(1-e2)]/5 = a2e2/5
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 (x2/a2)+(y2/b2) = 1 is 3, then the locus of the point is
(a) 3y2 – 2xy – 3b2 = 0 (b) 3y2+2xy – 3b2 = 0 (c) y2+xy – 3b2 = 0 (d) 2y2+xy – b2 = 0
(a) Any tangent to the ellipse is y = mx+√(a2m2+b2)
Let it pass through (α, β)
(β – mα)2 = a2m2+b2
⇒ m2(α2 – a2)-2mαβ+β2 – b2 = 0
m1m2 are the slopes
m1+m2 = 2αβ/(α2 – a2)
m1m2 = (β2 – b2)/(α2 – a2)
(1/m1)+(1/m2) = 3 ⇒ (m1+m2)/m1m2 = 3
⇒ [2αβ/(α2 – a2)]/[(β2 – b2)/(α2 – a2)]
⇒ 2αβ/(β2 – b2) ⇒ 3β2 – 3b2 = 2αβ
Locus of (α, β) is 0 3y2 – 2xy – 3b2 = 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 – af2 – bg2 – ch2 = 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 = c2 be the rectangular hyperbola, and let P (x1, y1) be a point on it.
Let Q(h, k) be the mid-point of PN. Then the coordinates of Q are (x1 , y1/2) .
∴ x1 = h and y1/2 = k
⇒ x1 = h and y1 = 2k
But (x1, y1) lies on xy = c2.
∴ h (2k) = c2 ⇒ hk = c2/2
Hence, the locus of (h, k) is xy = c2/2, which is hyperbola
19) The locus of the mid-points of the chords of the circle x2+y2 = 16 which are tangent to the hyperbola 9x2 – 6y2 = 144 is
(a) (x2+y2)2 = 16x2 – 9y2 (b) (x2+y2)2 = 9x2 – 16y2
(c) (x2 – y2)2 = 16x2 – 9y2 (d) none of these
(a) The given hyperbola is (x2/16) – (y2/9) …(i)
Any tangent to (i) is y = mx+√(16m2 – 9) …(ii)
Let (x1,y1) be the mid-point of the chord of the circle x2+y2 = 16
Then equation of the chord is T = S1
i.e. xx1+yy1 – (x12+y12) = 0 …(iii)
Since (ii) and (iii) represent the same line.
∴ m/x1 = -1/y1 = √(16m2-9)/-(x12+y12)
⇒ m = -(x1/y1) and (x12+y12)2 = y12(16m2-9)
⇒ (x12+y12)2 = 16×(x12/y12)y12 – 9y12 = 16x12 – 9y12
∴ locus of (x1,y1) is (x2+y2)2 = 16x2 – 9y2
20) The slope of the common tangents to the hyperbola (x2/92) – (y2/162) = 1 and (y2/92) – (x2/162) = 1
(a) -2,2 (b) -1,1 (c) 1,2 (d) 2,1
(b) Given hyperbolas are (x2/9)-(y2/16) = 1 …(i)
and (y2/16)-(x2/16) = 1 …(ii)
Any tangent to (i) y = mx±√(9m2-16) …(iii)
Putting in (ii), we get 16[mx±√(9m2-16)]2 – 9x2
⇒ (16m2 – 9)x2 ± 32m(√(9m2-16))x+144m2-256-144 = 0
⇒ (16m2 – 9)x2 ± 32m(√(9m2-16))x+(144m2 – 400) …(iv)
If (iii) is a tangent to (ii), then the roots of (iv) are real and equal.
∴ Disc. = 0
32×32m2(9m2-16) = 4(16m2 – 9)(144m2 – 400) = 64(16m2 – 9)(9m2-25)
⇒ 16m2(9m2-16) = (16m2 – 9)(9m2-25)
⇒ 144m4 – 256m2 = 144m4 – 481m2+225
⇒ 225m2 = 225
⇒ m2 = 1 ⇒ m = ±1
