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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Find the equation of the parabola with focus F(0,-3) and directrix y=3. |
| Answer» Correct Answer - `x^(2)-12y` | |
| 52. |
Find the equation of the parabola with vertex at the origin and focus F(0,5). |
| Answer» Correct Answer - `x^(2)=20y` | |
| 53. |
A line through the origin intersects the parabola `5y=2x^(2)-9x+10` at two points whose x-coordinates add up to 17. Then the slope of the line is __________ . |
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Answer» Correct Answer - 5 (5) Let the line be y=mx (1) solving it with `5y=2x^(2)-9x+10`, we get `or2x^(2)-(9+5m)x+10=0` Sum of roots `=(9+5m)/(2)=17` `or9+5m=34` `or5m=25` `orm=5` |
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| 54. |
A line L passing through the focus of the parabola `(y-2)^(2)=4(x+1)` intersects the two distinct point. If m be the slope of the line I,, thenA. `min(-oo, -1)uu(1, oo)`B. `m in (-oo, 0)uu(0, oo)`C. `min(-oo, 0)uu(0, oo)`D. none of these |
| Answer» Correct Answer - C | |
| 55. |
Let S be the focus of the parabola `y^2=8x` and let PQ be the common chord of the circle `x^2+y^2-2x-4y=0` and the given parabola. The area of the triangle PQS is -A. 4B. 3C. 2D. 8 |
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Answer» Correct Answer - A The parametric equations of the parabola are `x=2t^(2), y=4t`, Putting `x=2t^(2)` and y=4t in `x^(2)=y^(2)-2x-4y=0,` we get `4t^(4)+156t^(2)-4t^(2)-16t=0` `rArr" "4t^(2)+12t^(2)-16t=0` `rArr" "4t(t^(3)+3t-4)=0` `rArr" "t(t-1)(t^(2)+t+4)=0rArrt=0, t=1" "[because t^(2)+t+4ne0]` Thus, the coordinates of points of intersection of the circle and the parabola are O(0, 0) and P(2, 4). Clearly, these are diametrically opposite points on the circle. The coordinates of the focus S of the parabola are (2, 0) which lies on the circle. So, OSP is a right trangle. `:." Area of "DeltaOSP=1/2xxOSxxSP=1/2xx2xx4=4` |
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| 56. |
Two parabola have the same focus. If their directrices are the x-axis and the y-axis respectively, then the slope of their common chord is :A. `pm1`B. `4//3`C. `3//4`D. none of these |
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Answer» Correct Answer - A (1) Let the focus be (a,b). The equation are `S_(1):(x-a)^(2)+(y-b)^(2)=x^(2)` `S_(2):(x-a^(2))+(y-b)^(2)=y^(2)` Common chord `S_(1)-S_(2)=0` gives `x^(2)-y^(2)=0` `ory=pmx` |
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| 57. |
The value(s) of a for which two curves `y=ax^(2)+ax+(1)/(24)andx=ay^(2)+ay+(1)/(24)` touch each other is/areA. `(2)/(3)`B. `(1)/(3)`C. `(3)/(2)`D. `(1)/(2)` |
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Answer» Correct Answer - A::C 1,3 The two curves are symmetric about the line y=x. Hence, they touch each other on y=x. So, the point of contact is `(alpha,alpha)`. From any of the equations, we get `alpha=aalpha^(2)+aalpha+(1)/(24)` `or24aalpha^(2)+24alpha(a-1)+1=0` This equation should have Identical roots. `rArrD=0` `rArr(24)^(2)(a-1)^(2)-4(24a)=0` `rArr6a^(2)-13a+6=0` `rArr(2a-3)(3a-2)=0` `rArra=3//2,2//3` |
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| 58. |
If `A B`is a focal chord of `x^2-2x+y-2=0`whose focus is `S`and `A S=l_1,`then find `B Sdot` |
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Answer» `x^(2)-2x+y-2=0` `orx^(2)-2x+1=3-y` `or(x-1)^(2)=-(y-3)` The length of its latus rectum is 1 unit. Since AS, 1/2, and BS are in HP, we have `(1)/(2)=(2ASxxBS)/(AS+BS)` `or" "BS=(l_(1))/((4l_(1)-1))` |
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| 59. |
If `P(t^2,26),t in [0,2]`, is an arbitrary point on the parabola `y^2=4x ,Q`is the foot of perpendicular from focus `S`on the tangent at `P ,`then the maximum area of ` P Q S`is1 (b) 2(c) `5/(16)`(d) 5A. 1B. 2C. `5//16`D. 5 |
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Answer» Correct Answer - D (4) The equation of tangent at `P(r^(2),2t)" is "ty=x+t^(2)` It intersects the line x=0 at Q(0,t). Therefore, `"Area of "DeltaPQS=(1)/(2)|(0,t,1),(1,0,1),(t^(2),2t,1)|` `=(1)/(2)(t+t^(3))` Now, `(dA)/(dt)=(1)/(2)(1+3t^(2))gt0AAtinR` Therefore, the area is maximum for t=2. Hence, Maximum area `=(1)/(2)(2+8)=5` sq. units |
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| 60. |
The tangent to the parabola `y^(2)=4ax` at `P(at_(1)^(2), 2at_(1))" and Q"(at_(2)^(2), 2at_(2))` intersect on its axis, themA. `t_(1)=t_(2)`B. `t_(1)=-t_(2)`C. `t_(1)t_(2)=2`D. `t_(1)t_(2)=-1` |
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Answer» Correct Answer - B Tangents at P and Q intersect at `R(at_(1)t_(2),a(t_(1)+t_(2)))`. Since R lies at the axis of the parabola i.e. x-axis `:." "a(t_(1)+t_(2))=0rArrt_(2)=-t_(2)` |
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| 61. |
The normal drawn at a point `(at_(1)^(2), 2at_(1))` of the parabola `y^(2)=4ax`meets on the point `(ar_(2)^(2), 2at_(2))` thenA. `t_(1)=2t_(2)`B. `t_(1)^(2)=2t_(2)`C. `t_(1)t_(2)=1`D. none of these |
| Answer» Correct Answer - D | |
| 62. |
If `P(at_(1)^(2), 2at_(1))" and Q(at_(2)^(2), 2at_(2))` are two points on the parabola at `y^(2)=4ax`, then that area of the triangle formed by the tangents at P and Q and the chord PQ, isA. `1/2a^(2)|t_(1)-t_(2)|^(3)`B. `1/2a^(2)|t_(1)-t_(2)|^(2)`C. `a^(2)|t_(1)-t_(2)|^(3)`D. none of these |
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Answer» Correct Answer - A Let R be the point of intersection of tangents at P and Q. Then, the coordinates of R are `(at_(1), t_(2), a(t_(1)+t_(2)))`. `:." Area of "Delta PQR=1/2|Delta|`, where `Delta=|{:(at_(1)^(2),2at_(1), 1),(at_(1)^(2),2at_(2),1),(at_(1)t_(2),a(t_(1)+t_(2)),1):}|` `rArrDelta=a^(2)|{:(t_(1)^(2),2t_(1),1),(t_(2)^(2),2t_(2),1),(t_(1)t_(2),t_(1)+t_(2),1):}|` `rArrDelta=a^(2)|{:(t_(1)^(2)-t_(1)t_(2),t_(1)-t_(2),0),(t_(2)^(2)-t_(1)t_(1),t_(2)-t_(1),0),(t_(1)t_(2),t_(1)+t_(2),1):}|[{:("Appliyin",R_(1)toR_(1)-R_(3)),(,R_(2)toR_(2)-R_(3)):}]` `rArrDelta=a^(2)|{:(t_(1)(t_(1)-t_(2)),t_(1)-t_(2),0),(-t_(2)(t_(1)-t_(2)),-(t_(1)-t_(2)),0),(t_(1)t_(2),t_(1)+t_(2),1):}|` `rArrDelta=a^(2)(t_(1)-t_(2))^(2)|{:(t_(1),1,0),(-t_(2),-1,0),(t_(1)t_(2),t_(1)+t_(2),1):}|` `rArrDelta=a^(2)(t_(1)-t_(2))^(2)(-t_(1)+t_(2))=-a^(2)(t_(1)-t_(2))^(3)` Hence, area of `Delta PQR=1/2a^(2)|t_(1)-t_(2)|^(3)` |
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| 63. |
The equation of the parabola with its vertex at (1, 1) and focus at (3, 1) isA. `(x-3)^(2)=8(y-1)`B. `(y-1)^(2)=8(x-1)`C. `(y-1)^(2)=8(x-3)`D. `(x-1)^(2)=8(y-1)` |
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Answer» Correct Answer - B The vetex and focus of the parabola are (1, 1) and (3, 1) respectively. Clearly, axis is parallel to x-axis and Latusrectum = 4 (Distance between vertex and focus) `rArr" Latusrectum "=4xx2=8` Also, the vertex is on the left hand side of the focus. So, the parabola opens in the poitive direction of x-axis. Hence, the equation of the parabola is `(y-1)^(2)=8(x-1)` |
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| 64. |
Let `A (0,2),B` and C be points on parabola `y^(2)+x +4` such that `/_CBA (pi)/(2)`. Then the range of ordinate of C isA. `(-oo,0)uu (4,oo)`B. `(-oo,0] uu[4,oo)`C. `[0,4]`D. `(-oo,0)uu [4,oo)` |
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Answer» Correct Answer - B `A(0,2), B= (t_(1)^(2)-4,t_(1)), C =(t^(2) -4,t)` `/_CBA = (pi)/(2)` `rArr (2-t_(1))/(4-t_(1)^(2)). (t_(1)-t)/(t_(1)^(2)-t^(2)) =-1` `rArr (1)/(2+t_(1)). (1)/(t+t_(1)) =-1` `rArr t_(1)^(2) + (2+t) t_(1) + (2t+1) =0` For real `t_(1), (2+t)^(2) -4(2t-1) = t^(2) - 4t ge 0` `rArr t in (-oo,0] uu[4,oo)` |
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| 65. |
If the vertex of the parabola `y=x^(2)-8x+c` lies on x-axis, then the value of c, isA. -16B. -4C. 4D. 16 |
| Answer» Correct Answer - D | |
| 66. |
If a focal chord of `y^(2)=4ax` makes an angle `alphain[pi//4,pi//2]` with the positive direction of the x-axis, then find the maximum length of this focal shord. |
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Answer» Correct Answer - 8a units The length of focal chord making an angle `alpha` with the x-axis is `4a" cosec"^(2)alpha`. For `alphain[pi//4,pi//2]`, its maximum length is `4axx2=8a` units. |
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| 67. |
If the line passing through the focus `S`of the parabola `y=a x^2+b x+c`meets the parabola at `Pa n dQ`and if `S P=4`and `S Q=6`, then find the value of `adot` |
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Answer» Correct Answer - `pm(5)/(48)` The length of latus rectum of`y=ax^(2)+bx+c" is "1//|a|`. Now, the semi-latus rectum is the HM of SP and SQ. Then, we have `(1)/(SP)+(1)/(SQ)=(2)/(1//|2a|)` `or4|a|=(1)/(4)+(1)/(6)=(5)/(12)` `ora=pm(5)/(48)` |
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| 68. |
Show that the locus of a point that divides a chord of slope 2 of the parabola `y^2=4x` internally in the ratio `1:2` is parabola. Find the vertex of this parabola. |
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Answer» Correct Answer - Vertex `-=((2)/(9),(8)/(9))` Slope of the chord joining point `P(t_(1))andQ(t_(2))` on the parabola `y^(2)=4x` is `(2)/(t_(2)+t_(1))=2` (given) `:." "t_(2)+t_(1)=1` (1) Let point R(h,k) divide PQ in the ratio 1 : 2. `So," "h=(t_(2)^(2)+2t_(1)^(2))/(1+2)andk=(2t_(2)+2(2t_(1)))/(1+2)` `rArr" "3h=t_(2)^(2)+2t_(1)^(2)` (2) `and" "t_(2)+2t_(1)=(3k)//2` (3) From (1) and (3), we get `t_(1)=(3)/(2)k-1andt_(2)=2-(3)/(2)k` Putting these values in (2), we get `(2-(3)/(2)k)^(2)+2((3)/(2)k-1)^(2)=3h` Hence, locus of the point R is `(y-(8)/(9))^(2)=(4)/(9)(x-(2)/(9))`, which is parabola having vertex at `((2)/(9),(8)/(9))`. |
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| 69. |
If `y_(1),y_(2)` are the ordinates of two points P and Q on the parabola and `y_(3)` is the ordinate of the intersection of tangents at P and Q, thenA. `y_(1),y_(2),y_(3)" are in AP"`B. `y_(1),y_(3),y_(2)" are in AP"`C. `y_(1),y_(2),y_(3)" are in GP"`D. `y_(1),y_(3),y_(2)" are in GP"` |
| Answer» Correct Answer - B | |
| 70. |
If the line` x + y = 1 `touches the parabola `y^2-y + x = 0`, then the coordinates of the point of contact are:A. (1, 1)B. (1/2, 1/2)C. (0, 1)D. (1, 0) |
| Answer» Correct Answer - C | |
| 71. |
The angle between the tangents drawn from the point (-a, 2a) to `y^2`=4ax isA. `pi//4`B. `pi//2`C. `pi//3`D. `pi//6` |
| Answer» Correct Answer - B | |
| 72. |
The vertex of a parabola is the point (a,b) and latusrectum is of length `l`. If the axis of the parabola is along the positive direction of y-axis, then its equation is :A. `(x+a)^(2)=1/2(2 y-2b)`B. `(x-a)^(2)=1/2(2 y-2b)`C. `(x+a)^(2)=1/4(2 y-2b)`D. `(x-a)^(2)=1/8(2 y-2b)` |
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Answer» Correct Answer - B The equation of the referred to its vertex as the origin, axis along y-axis and latusrectum of length I is `x^(2)=Iy` Now, shifting the vertex at (a, b), the above equation reduces to `(x-a)^(2)=I(y-b)rArr(x-a)^(2)1/2(2y-ab)` |
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| 73. |
The locus of the vertices of the family of parabolas `y =[a^3x^2]/3 + [a^2x]/2 -2a` is:A. `xy=(105)/(64)`B. `xy=3/4`C. `xy=35/16`D. `xy=64/105` |
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Answer» Correct Answer - A We know that the vertex of the parabola given by `y=ax^(2)+bx+c` is at `((-b)/(2a),(-D)/(4a))`, So the coordinates of the verteds of the parabola `y=a^(3)/3x^(2)+a^(2)/2x-2a," are "((-3)/(4a),(-35a)/(16))` Thus, if (h, k) are the coordinates of the vertex. Then, `h=(-3)/(4a)" and K"=(-35a)/(16)rArrhk=(105)/(64)` Hence, the locus of (h, k) is `xy=(105)/(64)`. |
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| 74. |
Let a, r, s, t be non-zero real numbers. Let `P(at^(2),2at),Q(ar^(2),2ar)andS(as^(2),2as)` be distinct points on the parabola `y^(2)=4ax`. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K the point (2a,0). If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate isA. `((t+1)^(2))/(2t^(3))`B. `(a(t+1)^(2))/(2t^(3))`C. `(a(t^(2)+1)^(2))/(t^(3))`D. `(a(t^(2)+2)^(2))/(t^(3))` |
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Answer» Correct Answer - B The equations of the tangent and normal to `y^(2)=4ax" at "P(at^(2), 2at)" and S"(as^(2), 2as)` are respectively. `ty=x+at^(2)" and "y+sx=2as+as^(2)` or, `y-x/t=at" adn "y=+1/txx=(2a)/t+a/t^(3)` Adding these two equation, we get `2y=at+(2a)/t+a/t^(3)rArry=(a(t^(2)+1)^(2))/(at^(3))` |
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| 75. |
Find the equation of the parabola which is symmetric about the y-axis, and passes through the point `(2,-3)`. |
| Answer» Correct Answer - `3x^(2)-4y` | |
| 76. |
Statement 1 : The curve `y= - x^2/2 + x + 1` is symmetric with respect to the line `x=1`. Statement 2: A parabola is symmetric about its axis (A) Both 1 and 2 are true and 2 is the correct explanation of 1 (B) Both 1 and 2 are true and 2 is not a correct explanation of 1 (C) 1 is true but 2 is false (D) 1 is false but 2 is trueA. Statement I is correct, Statement II is correct, Statement II is not a correct explanation for statement IB. Statement I is correct, Statement II is incorrectC. Statement I is incorrect, Statement II is correctD. |
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Answer» Correct Answer - A `y = - (x^(2))/(2) + x + 1 implies y - (3)/(2) = - (1)/(2) (x - 1)^(2)` `implies` It is symmetric about x = 1 Hence, option (a) is correct. |
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| 77. |
If the distance of the point `(alpha,2)`from its chord of contact w.r.t. the parabola `y^2=4x`is 4, then find the value of `alphadot` |
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Answer» Correct Answer - `1-2sqrt(2)` The chord of contact of parabola `y^(2)=4x` w.r.t. point P(h,2) is `2y=2(x+h)` `or" "x-y+h=0` Distance of this chord from point P is 4. `:." "(|h-2+h|)/(sqrt(2))=4` `rArr" "|h-1|=2sqrt(2)` `rArr" "h=1pm2sqrt(2)` `rArr" "h=1-2sqrt(2)" as for "h=1+2sqrt(2)` point lies inside parabola) |
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| 78. |
Find the length of the common chord of the parabola `x^2=4(x+3)`and the circle `x^2+y^2+4x=0`. |
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Answer» Correct Answer - 4 Solving given equations, we have `-x^(2)-4x=4x+12` `rArr" "x^(2)+8x+12=0` `rArr" "x=-2,(x=-6` is not possible) From equation of parabola, `y^(2)=4(-2+3)=4` `:." "y=pm2`. So, curves intersect at (-2,2) and (-2,-2). So, length of common chord is 4. |
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| 79. |
The equation of the latus rectum of a parabola is `x+y=8`and the equation of the tangent at the vertex is `x+y=12.`Then find the length of the latus rectum. |
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Answer» Correct Answer - `8sqrt(2)` Since the equation of latus rectum and the equation of tangent both are parallel, and the distance them is one - fourth of the latus rectum, the distance between them is `a=|(-8+12)/(sqrt(1^(2)+1^(2)))|=(4)/(sqrt(2))=2sqrt(2)` `:." Length of latus rectum"=4a=4(2sqrt(2))=8sqrt(2)` |
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| 80. |
Match the conditions/expression in Column I With statement in Column II. Normals at P,Q,R are drawn to `y^(2) = 4x` which intersect at (3,0). Then, `{:("column I","column II"),(A."Area of" Delta PQR,p.2),(B. "Radius of circumcircle of" Delta PQR,q.(5)/(2)),(C. "Centroid of" Delta PQR,r. ((5)/(2),0)),(D. "Circumcentre of" Delta PQR,s. ((2)/(3),0)):}` |
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Answer» Correct Answer - `A to p; B to q: C to s; D to r` since, equation of normal to the parabola `y^(2) = 4ax` is `y + xt = 2at + at^(3)` passes through (3,0) `implies 3t = 2t + t^(2)` `implies t = 0, 1 -1` `:. ` Coordinates of the normals are `P(1,2), Q(0,0) R(1,-2)` Thus, A. Area of `Delta PQR = (1)/(2) xx 1 xx 4 = 2` C. Centriod of `Delta PQR = ((2)/(3), 0)` Equation of circle passing through P,Q,R is (x - 1)(x -1) + (y - 2) (y + 2) + `lambda` (x - 1) = 0 `implies 1 - 4 - lambda = 0` `implies lambda = - 3` `:.` Required equation of circle is `x^(2) + y^(2) - 5x = 0` `:.` Centre `((5)/(2) , 0)` and radius `(5)/(2)`. |
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| 81. |
Consider the parabola `y^2 = 8x.` Let `Delta_1` be the area of the triangle formed by the end points of its latus rectum and the point P(`1/2`,2) on the parabola and `Delta_2` be the area of the triangle formed by drawing tangents at P and at the end points of latus rectum. `Delta_1/Delta_2` is :A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B We know that the area of the trangle formed by three points on a prabola is twice the area of the triangle formed by tangents at those points (see illustration 3 on page 24.15). `:." "Delta_(1)=2Delta_(2)rArrDelta_(1)/Delta_(2)=2` |
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| 82. |
The locus of the midpoint of the focal distance of a variable pointmoving on theparabola `y^2=4a x`is a parabola whoselatus rectum is half the latus rectum of the original parabolavertex is `(a/2,0)`directrix is y-axis.focus has coordinates (a, 0)A. latus rectum is half the latus rectum of the original parabolaB. vertex is (a/2,0)C. directrix is y-axisD. focus has coordinates (a,0) |
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Answer» Correct Answer - A::B::C::D 1,2,3,4 Any point on the parabola is `P(at^(2),2at)`. Therefore, the midpoint of S(a,0) and `P(at^(2),2at)` is `R((a+at^(2))/(2),at)-=(h,k)` `:.h=(a+at^(2))/(2),k=at` Eliminate t, i.e., `2x=a(1+(y^(2))/(a^(2)))=a+(y^(2))/(a)` `i.e.," " 2ax=a^(2)+y^(2)` `i.e," "y^(2)=2a(x-(a)/(2))` It is a parabola with vertex at (a/2,0) and latus rectum 2a. The directrix is `x-(a)/(2)=-(a)/(2)` `i.e," "x=0` The focus is `x-(a)/(2)=(a)/(2)` i.e, x=a i.e., (a,0) |
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| 83. |
Find the equations of normal to the parabola `y^2=4a x`at the ends of the latus rectum. |
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Answer» End point of latus rectum of parabola `y^(2)=4ax` are P(a,2a) and Q(a-2a). Comparing with `(at^(2),2at)` , we get t=1 for point P and t=-1 for point Q. Now, equation of normal at point `(at^(2),2at)` is y=-tx+2at+ `at^(3)` So, equation of normal at point P is y=-x+2a+a or x+y=3a. Also, equation of normal at point Q is y=x-2a-a or x-y=3a. |
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| 84. |
The point of intersection of the tangents at the ends of the latusrectum of the parabola `y^2=4x`is_____________ |
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Answer» Correct Answer - (-1,0) The coordinates of extremities of the latusrectum of `y^(2) = 4x` are (1,2) and (1,-2) Equation of tangents at these points are `y. 2 = (4(x + 1))/(2) implies 2y = 2 (x + 1)` and `y (-2) = (4(x + 1))/(2)` `implies -2y = 2 (x + 1)` `:. -2 (x + 1) = 2 (x + 1)` `implies 0 = 4 (x + 1)` `implies - 1 = x implies y = 0` Therefore, the required point is (-1,0) |
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| 85. |
Let PQ be the latus rectum of the parabola `y^2 = 4x` with vetex A. Minimum length of the projection of PQ on a tangent drawn in portion of Parabola PAQ isA. `sqrt(2)a`B. `2asqrt(2)`C. `2a`D. `3asqrt(2)` |
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Answer» Correct Answer - B Let tangent at `P(at^(2),2at)` make an angle `theta` with x-axis. Then `tan theta = (1)/(t)` Projection of BC on tangent `= BC sin theta = (4a)/(sqrt(1+t^(2))) ge 2a sqrt(2) (as -1 le t le 1)` |
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| 86. |
The equations of the common tangents to the parabola `y = x^2 and y=-(x-2)^2` is/are :A. y=4(x-1)B. y=0C. y=-4(x-1)D. y=-30x-50 |
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Answer» Correct Answer - A::B 1,2 If y=mx+c is tangent to `y=x^(2)`, then `x^(2)-mx-c=0` has equal root. So, `m^(2)+4c=0` `orc=-(m^(2))/(4)` So, the tangent to `y=x^(2)` is `y=mx-(m^(2))/(4)` Since this is also tangent to `y=-(x-2)^(2)`, `mx-(m^(2))/(4)=-x^(2)+4x-4` has equal roots. So, `x^(2)+(m-4)x+(4-(m^(2))/(4))=0` has equal roots. So, `(m-4)^(2)-4(4-(m^(2))/(4))=0` `orm^(2)=8m+16+m^(2)-16=0` `orm=0,4` So,, y=0 and y=4x-4 is the tangent. |
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| 87. |
The equation of the common tangents to the parabola y = x2 and y = –(x – 2)2 is / are (a) y = 4(x–1)(b) y = 0 (c) y = –4(x–1) (d) y = –30x–50 |
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Answer» Correct option (a),(b) Explanation: Let y = mx + c is tangent to y = x2 mx + c = x2 ⇒ x2– mx – c = 0 has equal roots m2 + 4c = 0 y = mx – m2/4 is tangent to y = –(x–2)2 also mx – 4 m2 = –x2 + 4x – 4 x2 +(m – 4)x + 4 - m2/4 = has equal roots (m – 4)2 - 4(4 - m2/4) = 0 m2 + 16 – 8m – 16 + m2 = 0 m2 – 4m = 0 m = 0, 4 Equation of tangent are y = 0 and y = 4x – 4 |
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| 88. |
The curves `x^(2) +y^(2) +6x - 24y +72 = 0` and `x^(2) - y^(2) +6x +16y - 46 = 0` intersect in four points P,Q,R and S lying on a parabola. Let A be the focus of the parabola, thenA. `AP + AQ +AR +AS = 20`B. `AP + AQ +AR +AS = 40`C. vertex of the parabola is at `(-3,1)`D. coordinates of A are `(-3,1)` |
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Answer» Correct Answer - B::C The points of intersection P,Q,R,S lie on `2x^(2) +12x -8y +26 =0` (by adding equations) i.e., `(x+3)^(2) =4(y-1)` So, vertex is `(-3,1)` and focus is `(-3,2)`. Points of intersection also lie on `y^(2) - 20y +59 =0`. (by substracting equations) `:.` Sum of two of the ordinates `= 20` `:. AP + AQ + AR + AS = 40` |
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| 89. |
If two of the three feet of normals drawn from a point to the parabola `y^2=4x`are (1, 2) and `(1,-2),`then find the third foot. |
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Answer» The sum of the ordinates of the feet is `y_(1)+y_(2)+y_(3)=0`. Therefore, `2+(-2)+y_(3)=0` `or" "y_(3)=0` So, the third foot is (0,0). |
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| 90. |
Find the equation of normal to the parabola `y=x^2-x-1`which has equal intercept on the axes. Also find the point where this normalmeets the curve again. |
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Answer» Here, given equation of parabola is not standard. So, we use differentiation method to find the equation of normal. Since normal has equal intercepts on axes, its slope is -1 Now, differentiating equation `y=x^(2)-x-1` on both sides w.r.t. x, we get `(dy)/(dx)=2x-1`. This is the slope of tangent to the curve at any point on it. Thus, slope of normal at any point on it is, `(dx)/(dy)=(1)/(1-2x)=-` (given) `:." "x=1` Putting x=1 in the equation of curve, we get y=-1. So, equation of normal at point (1,-1) on the curve is `y-(-1)=-1(x-1)` `orx+y=0` Solving this equation of normal with the equation of parabola, we have `-x=x^(2)-x-1` `orx^(2)=1` `orx=pm1`. Hence, normal meet parabola again at point whose abscissa is -1. Putting x=-1 in the equation of curve, we get y=1. So, normal meets the parabola again at (-1,1). |
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| 91. |
IF three distinct normals to the parabola `y^(2)-2y=4x-9` meet at point (h,k), then prove that `hgt4`. |
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Answer» Given parabola is `(y-1)^(2)=4(x-2)`. Equation of normal to parabola having slope m is `y-1=m(x-2)-2m-m^(3)` It passes through (h,k) `:." "k-1=m(h-2)-2m-m^(3)` `or" "m^(3)+(4-h)m+k-1=0` This equation has three distinct real roots. `:." "3m^(2)+(4-h)=0` has two distinct real roots. `:." "4-hlt0orhgt4` |
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| 92. |
From the point (15, 12) three normals ae drawn to the parabola y2 = 4x, then centroid of triangle formed by three co-normal points is (a) (5, 0) (b) (5, 4) (c) (9, 0) (d) (26/3 , 0) |
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Answer» Correct option (d) (26/3, 0) Explanation: Let equation of normal be y = –tx + 2t + t3 It passes through (15, 12). So 12 = –15t + 2t + t3 t3 –13t –12 = 0 (t + 1) ( t + 3) (t – 4) = 0 t = –1, –3, 4 Points are (at2 , 2at) i.e. (1, – 2), (9, – 6), (16, 8) Centroid is (1 + 9 + 16/3 , -2 - 6 + 8/3) = (26/3, 0) |
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| 93. |
Line y=2x-b cuts the parabola `y=x^(2)-4x` at points A and B. Then the value of b for which `angleAOB` is a right is (where O is origin) _________ . |
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Answer» Correct Answer - 7 (7) The line is y=2x-b,i.e., `1=(2x-y)/(b)` Homogenizing parabola with line, we get `x^(2)-4x((2x-y)/(b))-y((2x-y)/(b))=0` Since `angleAOB=90^(@)` Coefficient of `x^(2)+"coefficient of "y^(2)=0` `or1-(8)/(b)+(1)/(b)=0` `orb=7` |
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| 94. |
Find the number of distinct normals that can be drawn from `(-2,1)`to the parabola `y^2-4x-2y-3=0`A. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - A The equation of the parabola is `(y-1)^(2)=4(x+1)`. The equation of any normal to this parabola is `y-1=m(x+1)-2m-m^(3)` If passes through (-2, 1). Then, `0=-m-2m-m^(3)rArrm^(3)+3m=0rArrm=0" "[becausem^(2)+3ne0]` So, there is only one normal passing through (-2, 1). |
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| 95. |
The slopes of tangents drawn from a point `(4, 10)` to parabola `y^2=9x` areA. `"1/4, 3/4"`B. `"1/4, 9/4"`C. `"1/4, 1/3"`D. none of these |
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Answer» Correct Answer - B The equation of a tangent of slope m to the parabola `y^(2)=9x` is `y=mx+9/(4m)` If it passes through (4, 10), then `10=4m+9/(4m)rArr16m^(2)-40m+9=0rArrm=1/4, 9/4` |
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| 96. |
If the line x+y=6 is a normal to the parabola `y^(2)=8x` at point (a,b), then the value of a+b is ___________ . |
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Answer» Correct Answer - 6 (6) Slope of the line =-1 From the curve, `(dy)/(dx)=(4)/(y)` Hence, slope of normal `=-(y)/(4)=-1ory=4` Putting y=4 in the equation of curve, we have x=2. Hence, the point is (2,4). |
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| 97. |
The normal at the point `(bt_1^2, 2bt_1)` on the parabola `y^2 = 4bx` meets the parabola again in the point `(bt_2 ^2, 2bt_2,)` thenA. `t_(2)=t_(1)+2/t_(1)`B. `t_(2)=t_(1)-2/t_(1)`C. `t_(2)=-t_(1)+2/t_(1)`D. `t_(2)=t_(1)-2/t_(1)` |
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Answer» Correct Answer - B The equation of the normal to the parabola `y^(2)=4bx" at "P(bt_(1)^(2), 2bt_(1))` is `y+t_(1)xx=2bt_(1)+bt_(1)^(3)` If it passes thorugh `(bt_(2)^(2), 2bt_(2))`, then `2bt_(2)+bt_(1)t_(2)^(2)=2bt_(1)+bt_(1)^(3)` `rArr" "bt_(1)(t_(2)^(2)-t_(1)^(2))=2b(t_(1)-t_(2))` `rArr" "t_(1)(t_(2)+t_(1))=-2rArrt_(2)=-t_(1)-2/t_(1)` |
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| 98. |
The value of `thetain(-pi/2, pi/2)` for which the line `y=x" cos "theta+4" cos" ^(3)" "theta-14" cos "theta-1` is a normal to the parabola `y^(2)=16x,` isA. `pi//3`B. `pi//6`C. `pi//9`D. `pi//4` |
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Answer» Correct Answer - C The slope of the given line is m = cos `theta`. We know that the line y = mx + c is a normal to the parabola `y^(2)=4ax`, if `c=-2am-am^(3)`. Therefore, the given line will be a normal to the parabola `y^(2)=16x,` is `4 cos^(3)theta-14costheta-1=-8costheta-4cos^(2)theta` `rArr" "8cos^(3)theta-6costheta=1` `rArr" "2(4cos^(3)theta-3costheta)=1` `rArr" "2cos3theta=1rArrcos3theta=1/2rArr3theta=pi/3theta=pi/9` |
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| 99. |
If `(h ,k)`is a point on the axis of the parabola `2(x-1)^2+2(y-1)^2=(x+y+2)^2`from where three distinct normals can be drawn, then prove that `h > 2.`A. `hgt2`B. `hlt4`C. `hgt8`D. `hlt8` |
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Answer» Correct Answer - A We have, `2(x-1)^(2)+(y-1)^(2)=(x+y+2)^(2)` `rArr" "sqrt((x-1)^(2)+(y-1)^(2))=|(x+y+2)/(sqrt(1+1))|^(2)` Clearly, it represents a parabola having its focus at (1, 1) and directrix x + 2 = 0. The equation of the axis is `y-1(x-1) i.e. y = x` Semi-latusrectum = Length of `bot` from (1, 1) on the directrix `rArr" Semi-laturectum "=|(1+1+2)/(sqrt(1+1))|=2sqrt2`. The coordinates of the vertex are (0, 0). So, the equation of the axis in parametrix form if `(x-0)/("cos "pi//4)=(y-0)/("sin "pi//4)" ...(i)"` We know that three distinct normals can be drawn from a point (h, 0) on the axis of the parabola `y^(2)=4ax` if `hgt2a("= semi-latusretum")`. The coordinates of a point on the axis (i) at a distance `2 sqrt2` from the vertex are given by `x/("cos "pi//4)=y/("sin "pi//4)=2sqrt2rArrx=2, y=2`. Hence, three normals can be drawn from (h, k), if `hgt2`. |
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| 100. |
Normals at P, Q, R are drawn to `y^(2)=4x` which intersect at (3, 0). Then, are of `DeltaPQR`, isA. `(2//3,0)`B. `(2//5,0)`C. `(5//2,0)`D. none of these |
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Answer» Correct Answer - A From example 86, we obtain that the coordinates of P, Q and R are (0, 0), (1, 2), and (1, -2). `:." Coordinates of the centriod of "DeltaPQR" are "(23, 0)` ALITER The centriod of triangle formed by the feet of normal (Co-normal) drawn from a point (h, k) to the parabola `y^(2)=4ax` has the coordinates `(2/3(h, -2a), 0)`. In this case, we have h = 3, k=0 and a=1. So, coordinates of the centroid are (2/3, 0). |
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