Mathametics 05 – Coordinate geometry

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

Screenshot 20200810 113458 2

(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

Screenshot 20200810 113636 2

(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+a…(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

Screenshot 20200810 113752 2

(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)

Screenshot 20200810 114019 2

(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

⇒ m22 – 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

Screenshot 20200810 114145 2

(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

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