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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.
| 151. |
At what point on the parabola `y^2=4x`the normal makes equal angle with the axes?`(4,4)`(b) `(9,6)`(d) `(4,-4)`(d) `(1,+-2)`A. (4,4)B. (9,6)C. (4,-4)D. `(1,pm2)` |
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Answer» Correct Answer - D (4) We have a=1. Normal at `(m^(2),-2m)" is "y=mx-2-m^(3)` Given that normal makes equal angle with the axes, its slope is `m=pm1`. Therefore, point P is `(m^(2),-2m)-=(1,pm2)`. |
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| 152. |
The line `2x+y+lamda=0` is a normal to the parabola `y^(2)=-8x,` is `lamda`=A. 12B. -12C. 24D. -24 |
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Answer» Correct Answer - C (3) y=mx+c is a normal to `y^(2)=4ax` if `c=-2am-am^(3)`, `y=-2x-lamda` (Given normal) `:." "m=-2,a=-2` `or-lamda=-2an-am^(3)=-2(-2)(-2)-(-2)(-2)^(3)=-24` `orlamda=24` |
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| 153. |
If the normal chord of the parabola `y^(2)=4 x` makes an angle `45^(@)` with the axis of the parabola, then its length, isA. 8B. `8sqrt2`C. 4D. `4sqrt2` |
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Answer» Correct Answer - B The equation of a normal to `y^(2)=4x" at "P(t^(2), 2t)` is `y+tx=2t+t^(3)` This makes an angle of `45^(@)` with x-axis. Therefore, t=-1. If it culs the parabola at `Q(t_(1)_^(2),2t_(1))`, then `t_(1)=-t-2/t=3` Thus, the coordinates of P and Q are (1, -2) and (9, 0) respectively. `:." "PQ=sqrt(64+64)=8sqrt2` |
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| 154. |
Three normals are drawn from the point (7, 14) to the parabola `x^2-8x-16 y=0`. Find the coordinates of the feet of the normals. |
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Answer» The equation of the given parabola is `x^(2)-8x-16y=0` Differentiating w.r.t. x, we get `2x-8-16(dy)/(dx)=0or(dy)/(dx)=(x-4)/(8)` Let normal be drawn at point (h,k) on the curve. `:." "k=(h^(2)-8h)/(16)` Also, slope of the normal at `(h,k)=(8)/(4-h)` Therefore, equation of the normal at (h,k) is `y-kj=((8)/(4-h) )(x-h)` If the normal passes through (7,14), then we have `14-(h^(2)-8h)/(16)=((8)/(4-h))(7-h)` `rArr" "(4-h)(224-h^(2)+8h)=128(7-h)` `rArr" "h^(3)-12h^(2)-64h=0` `rArr" "h(h-16)(h+4)=0` `rArr" "h=0,-4,16` For h=0, k=0 for h=-4, k=3 and for h=16, k=8. Therefore, the coordinates of the feet of the normal are (0, 0), (-4,3) and (16,8). |
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| 155. |
The length of the chord of the parabola `x^2=4ay` passing through the vertex and having slope `tanalpha is `(a>0)`:A. 2 a cosec `alpha` cot `alpha`B. `4a tan alpha sec alpha`C. `4a cos alpha cot alpha`D. `4a sin alpha tan alpha` |
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Answer» Correct Answer - A Let A be the vertex and AP be a chord of `x^(2)=4"ay"` such that slope of AP is tan `alpha`. Let the coorddinates of P be `(2at, at^(2))` Then, `"Slope of AP"=(at^(2))/(2at)=t/2rArrtan alpha=t/2rArrt=2 tan alpha` `"Now, "` `APsqrt((2at-t)^(2)+(at^(2)-0)^(2))` `rArr" "AP=atsqrt(4+1^(2))=2a" tan"alphasqrt(4+4"tan"^(2)alpha)=4a tan alpha sec alpha` |
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| 156. |
Which of the following points lie on the parabola x2 = 4ay? A. x = at2, y = 2at B. x = 2at, y = at2 C. x = at2, y = at D. x = 2at, y = at |
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Answer» Option : (B) Given that, The equation of the parabola is x2 = 4ay. We know, The parametric equation of the point on the parabola is (2at, at2) |
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| 157. |
The vertex of the parabola `x^2 + y^2 - 2xy - 4x - 4y + 4 = 0` is atA. (1, 1)B. (-1, -1)C. `(1/2,1/2)`D. none of these |
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Answer» Correct Answer - C we have, `x^(2)+y^(2)-2xy-4x-4y+4=0` `rArr" "(x-y)^(2)=4(x+y-1)` `rArr" "{(x-y)/(sqrt(1^(2)+1^(2)))}^(2)=2sqrt2{(x+y-1)/(sqrt(1^(2)+1^(2)))}` `rArr" "(x-y)/(sqrt(1^(2)+1^(2)))=" Y and "(x+y-1)/(sqrt(1^(2)+1^(2)))=X` Then, equation (i) reduces to `y^(2)=2sqrt2X`. The coordinates of the vertex of this parabola are (X = 0, Y = 0). So, the coordinates of the vertex of the given parabola are given by `(x+y-1)/(sqrt(1^(2)+1^(2)))=" 0 and "(x-y)/(sqrt(1^(2)+1^(2)))=0` i.e. x+y-1=0 and x-y=0 `rArr" "x=y=1/2` Hence, the coordinates of the vertex are (1/2, 1/2). |
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| 158. |
The length of the latusrectum of the parabola `2{(x-a)^(2)+(y-a)^(2)}=(x+y)^(2),` isA. 2aB. `2sqrt(2)a`C. 4aD. `sqrt2a` |
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Answer» Correct Answer - B We have `2{(x-a)^(2)+(y-a)^(2)}=(x+y)^(2)` `rArr" "sqrt((x-a)^(2)+(y-a)^(2))=1/sqrt2|x+y|` `rArr" "sqrt((x-a)^(2)+(y-a)^(2))=|(x+y)/sqrt2|` Clerly, this equation represents a parabola having its focus at (a, a) and directrix x + y = 0 `:.` L.R.=2 (Distance between focus and directrix) `rArr" "L.R.=2|(a+a)/(sqrt(1+1))|=2sqrt2a` |
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| 159. |
The equation of the directrix of the parabola `25{(x-2)^(2)+(y+5)^(2)}=(3x+4y-1)^(2),` isA. 3x+4y=0B. 3x+4y+1=0C. 4x-3y=0D. 3x+4y+1=0 |
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Answer» Correct Answer - B We have, `25{(x-2)^(2)+(y+5)^(2)}=(3x+4y-1)^(2),` `rArr" "sqrt((x-2)^(2)+(y+5)^(2))=|(3x+4y-1)/(sqrt(3^(2)+4^(2)))|` `rArr" SP=PM"`, where $ (2, 5) , P(x, y) and PM is the length of perpendicular from P on the line 3x+4y-1=0 Hence, 3x+4y-1=0 is the directrix adn $ (2, -5) is the focus of the given parabola. |
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| 160. |
Find the equation of the tangent to the parabola `y^2=8x`having slope 2 and also find the point of contact. |
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Answer» The equation of the tnagent to `y^(2)=4ax` having slope m is `y=mx+(a)/(m)` Hence, for the given parabola, the equation of the tangent is `y=2x+(2)/(2)ory=2x+1` and the point of contact is `((a)/(m^(2)),(2a)/(m))-=((2)/(2^(2)),(2(2))/(2))-=((1)/(2),2)` |
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| 161. |
y=x is tangent to the parabola `y=ax^(2)+c`. If a=2, then the value of c isA. 1B. `-1//2`C. `1//2`D. `1//8` |
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Answer» Correct Answer - D (4) `y=ax^(2)+c` `:.(dy)/(dx)=2ax=1` Therefore, the point of contact of the tangent is `((1)/(2a),(1)/(4a)+c)` Since it lies on y=x, we have `c=(1)/(4a)` Thus, c=1/8 for a=2. |
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| 162. |
Tangent to the parabola `y=x^(2)+ax+1` at the point of intersection of the y-axis also touches the circle `x^(2)+y^(2)=r^(2)`. Also, no point of the parabola is below the x-axis. The radius of circle when a attains its maximum value isA. `1//sqrt(10)`B. `1//sqrt(5)`C. 1D. `sqrt(5)` |
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Answer» Correct Answer - B (2) Since no point of the parabola is below the x-axis, `D=a^(2)-4le0` Therefore, the maximum value of a is 2. The equation of the parabola when a=2 is `y=x^(2)+2x+1` It intersect the y-axis at (0,1). The equation of the tangent at (0,1). y=2x+1 Since y=2x+1 touches the circle `x^(2)+y^(2)=r^(2)`, we get `r=(1)/(sqrt(5))` |
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| 163. |
Tangent to the parabola `y=x^(2)+ax+1` at the point of intersection of the y-axis also touches the circle `x^(2)+y^(2)=r^(2)`. Also, no point of the parabola is below the x-axis. The slope of the tangents when the radius of the circle is maximum isA. -1B. 1C. 0D. 2 |
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Answer» Correct Answer - C (3) The equation of the tangent at (0,1) to the parabola `y=x^(2)+ax+1` is y-1=a(x-0) `orax-y+1=0` As it touches the circle, we get `r=(1)/(sqrt(a^(2)+1))` The radius is maximum when a=0. Therefore, the equation of the tangent is y=1. Therefore, the slope of the tangent is 0. |
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| 164. |
Tangent to the parabola `y=x^(2)+ax+1` at the point of intersection of the y-axis also touches the circle `x^(2)+y^(2)=r^(2)`. Also, no point of the parabola is below the x-axis. The minimum area bounded by the tangent and the coordinate axes isA. 1B. `1//3`C. `1//2`D. `1//4` |
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Answer» Correct Answer - D (4) The equation of tangent is y=ax+1. The intercepts are `-1//a` and 1. Therefore, the area of the triangle bounded by tangent and the axes is `(1)/(2)|-(1)/(a)*|=(1)/(2|a|)` It is minimum when a=2. Therefore, Minimum area `=(1)/(4)` |
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| 165. |
Find the equation of the normal to the parabola `y^2 = 4x` which is parallel to `y - 2x + 5 = 0`A. y=2x+12B. y=2x-12C. y=2x+8D. y=-2x+12 |
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Answer» Correct Answer - B The equation of the normal to the parabola `y^(2)=4ax" at "(am^(2), -2m)` is `y=mx-2am-am^(3)`, where m is the slope of the normal. Here, a = 1. So, the equation of the normal at `(m^(2), -2m)` is `y = mx-2m -m^(3)" ....(i)"` If the normal is parallel to the line y=2x-5. Then, m = Slope of the line `y = 2x - 5 rArr m=2` Putting the value of m in (i), we obtain y=2x-12 as the equation of the requaired normal. |
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| 166. |
Consider the two curves `C_1` ; `y^2 = 4x`, `C_2` : `x^2 + y^2 - 6x + 1 = 0` then :A. `C_(1)" and "C_(2)` touch each other at one pointB. `C_(1)" and "C_(2)` touch eacth other exactly at two pointC. `C_(1)" and "C_(2)` intersect ( but do not touch) at exactly two pointsD. `C_(1)" and "C_(2)` neither intersect not touch each other |
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Answer» Correct Answer - B The x-coordinates of the points of intersection of `C_(1)" and "C_(2)` are roots of the equation `x^(2)+4x-6x+2=0rArrx=1` Putting x = 1 in `y^(2)=4x`, we get `y = +- 2`. Thus, two curves intersect at (1, 2) and (1, -2). Equations of tangents to `C_(1)" and "C_(2)` at (1, 2) are `2y=2(x+1)" and "x+2y-3(x+1)+1=0` `"or, y=x+1 and y=x+1"` Thus, `C_(1)" and "C_(2)` touch each other at P(1, 2). Similarly, `C_(1)" and C_(2)` have the same tangent x + y + 1 = 0 at Q(1, -2). |
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| 167. |
y=x is tangent to the parabola `y=ax^(2)+c`. If (1,1) is the point of contact, then a isA. `1//4`B. `1//3`C. `1//2`D. `1//6` |
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Answer» Correct Answer - C (3) If (1,1) is point of contact, then a=1/2. |
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| 168. |
y=x is tangent to the parabola `y=ax^(2)+c`. If c=2, then the point of contact isA. (3,3)B. (2,2)C. (6,6)D. (4,4) |
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Answer» Correct Answer - D (4) If c=2, then the point of contact is `(1//2a,1//4a+2)`. Since it lies on the line y=x, we have `(1)/(2a),(1)/(4a)+2` `i.e," "a=(1)/(8)` Therefore, the point of contact is (4,4) |
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| 169. |
Parabola `y^2=4a(x-c_1) and x^2=4a(y-c_2)` where `c_1 and c_2` are variables, touch each other. Locus of their point of contact isA. `xy=2a^(2)`B. `xy=4a^(2)`C. `xy=a^(2)`D. none of these |
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Answer» Correct Answer - B (2) Let P(x,y) be the point of contact. At P, both of them must have the same slope. We have `2y(dy)/(dx)=4a,2x=4a(dy)/(dx)` Eliminating `dy//dx`, we get `xy=4a^(2)`. |
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| 170. |
Consider the two curves `C_1: y^2=4x ,C_2: x^2+y^2-6x+1=0`. Then`C_1`and `C_2`touch each other only at onepoint`C_1`and `C_2`touch each other exactly at twopoint`C_1`and `C_2`interesect (but do not touch) atexactly two points.`C_1`and `C_2`neither intersect nor touch eachother.A. Statement-1 is True, Statement - 2 is true, Statement-2 is a correct explanation for Statement-1`B. Statement-1 is True, Statement - 2 is true, Statement-2 is not a correct explanation for Statement-3C. Statement-1 is True, Statement - 2 is False.D. Statement-1 is True, Statement - 2 is True. |
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Answer» Correct Answer - B Statement-1 and 2 both are true. But statement-2 is not a correct explanation for statement-1. |
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| 171. |
Let A,B and C be three distinct points on `y^(2) = 8x` such that normals at these points are concurrent at P. The slope of AB is 2 and abscissa of centroid of `Delta ABC` is `(4)/(3)`. Which of the following is (are) correct?A. Area of `DeltaABC` is 8 sq. unitsB. Coordinates of `P -= (6,0)`C. Angle between normals are `45^(@),45^(@),90^(@)`D. Angle between normals are `30^(@),30^(@),60^(@)` |
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Answer» Correct Answer - A::B::C Let `A = (2t_(1)^(2),4t_(1)), B = (2t_(2)^(2), 4t_(2))` and `C = (2t_(3)^(2),4t_(3))` Slope of `AB = 2 rArr t_(1) + t_(2) =1` and `t_(1) + t_(2) + t_(3) =0` So, `t_(3) =-1` Also, `(2(t_(1)^(2)+t_(2)^(2)+t_(3)^(2)))/(3) =(4)/(3) rArr t_(1)^(2)+t_(2)^(2) =1` `rArr t_(1) =1, t_(2) =0` `A = (2,4), B =(0,0)` and `C = (2,-4)` Hence `P = (6,0)` |
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| 172. |
A line `ax +by +c = 0` through the point `A(-2,0)` intersects the curve `y^(2)=4a` in P and Q such that `(1)/(AP) +(1)/(AQ) =(1)/(4)` (P,Q are in 1st quadrant). The value of `sqrt(a^(2)+b^(2)+c^(2))` isA. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - B Let `P(-2+r cos theta, r sin theta)`. `rArr r^(2) sin^(2) theta - 4r cos theta +8 =0` `rArr (1)/(AP) + (1)/(AQ) =(1)/(r_(1)) +(1)/(r_(2)) =(1)/(4)` `rArr cos theta =(1)/(2) rArr tan theta = sqrt(3)` So, equation of line is `y -0 = sqrt(3) (x+2)` `rArr sqrt(3) x -y +2sqrt(3) =0` So `sqrt(a^(2)+b^(2)+c^(2)) =4`. |
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| 173. |
If the focus of a parabola is at (0,-3) and its directrix is y = 3, then its equation isA. `x^(2)=-12y`B. `x^(2)=12y`C. `y^(2)=-12x`D. `y^(2)=112x` |
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Answer» Correct Answer - A Let P (x, y) be any point on the parabola. Then, by difinition, we have `sqrt((x-0)^(2)+(y+3)^(2))=y-3rArrx^(2)=-12y` |
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| 174. |
If the normals to the parabola `y^2=4a x`at three points `(a p^2,2a p),`and `(a q^2,2a q)`are concurrent, then the common root of equations `P x^2+q x+r=0`and `a(b-c)x^2+b(c-a)x+c(a-b)=0`is`p`(b) `q`(c) `r`(d) `1`A. pB. qC. rD. 1 |
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Answer» Correct Answer - D (4) Normal at points `(ap^(2),2ap),(aq^(2),2aq),and(ar^(2),2ar)` are concurrent. Hence, the points are co-normal points. Therefore, p+qr=0 So, `px^(2)+qx+r=0` has one root which is x=1. Therefore, the common root is 1, which also satisfies the second equation. |
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| 175. |
`min[(x_1-x^2)^2+(3+sqrt(1-x1 2)-sqrt(4x_2))],AAx_1,x_2 in R ,`is`4sqrt(5)+1`(b) `3-2sqrt(2)``sqrt(5)+1`(d) `sqrt(5)-1`A. `4sqrt(1)`B. `3-2sqrt(2)`C. `sqrt(5)+1`D. `sqrt(5)-1` |
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Answer» Correct Answer - B (2) Let `y_(1)=3+sqrt(1-x_(1)^(2))andy_(2)=sqrt(4x_(2))` `orx_(1)^(2)+(y_(1)-3)^(2)=1andy_(2)^(2)=4x_(2)` Thus, `(x_(1),y_(1))` lies on the circle `x^(2)+(y-3)^(2)=1`. Also, `(x_(2),y_(2))` lies on the upper half of the parabola `y^(2)=4x`. Thus, the given expression is square of the shortest distance between the curves `x^(2)+(y-3)^(2)=1andy^(2)=4x`. Now, the shortest distance always occurs along the common normal to the curves and the normal to the circle passes through the center of the circle. Normal to the parabola `y^(2)=4x" is "y-mx-2m-m^(3)`. It passes through (0,3). Therefore, `m^(3)+2m+3=0`, which has only one root, m=-1. Hence, the required minimum distance is `sqrt(1+1)-1=sqrt(2)-1`. |
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| 176. |
Statement-1: Length of the common chord of the parabola`y^(2)=8x` and the circle `x^(2)+y^(2)=9` is less than the length of the latusrectum of the parabola. Statement-2: If vertex of a parabola lies at the point (a. 0) and the directrix is x + a = 0, then the focus of the parabola is at the point (2a, 0).A. Statement-1 is True, Statement - 2 is true, Statement-2 is a correct explanation for Statement-1`B. Statement-1 is True, Statement - 2 is true, Statement-2 is not a correct explanation for Statement-5C. Statement-1 is True, Statement - 2 is False.D. Statement-1 is True, Statement - 2 is True. |
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Answer» Correct Answer - C Parabola `y^(2)=-8x` and the circle `x^(2)=y^(2)=9` intersect at `P(1, 2sqrt2)" and "Q(1, -2sqrt2)`. `:. PQ=4sqrt2lt"Latusrectum (=8)"` So, statement-1m is true. Since vertex (a, 0) is the mid-point of the line segment joining the focus and (-a, 0). So, coordinates of forcus are (3a, 0). Hence, statement-2 is not true. |
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| 177. |
The equation of a parabola having focus (3, 0) and directrix x + 3 = 0, isA. `y^(2)=12x`B. `y^(2)=-12x`C. `x^(2)=12y`D. `x^(2)=-12y` |
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Answer» Correct Answer - A Let p (x, y) be any point on the parabola. Then, `sqrt((x-3)^(2)+(y+0)^(2))=|(x+3)/(sqrt(1+0))|" [Using SP= PM]"` `rArr" "(x-3)^(2)+y^(2)=(x+3)^(2)rArry^(2)=12x` |
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| 178. |
The locus of the poles of tangents to the parabola `y^(2)=4ax` with respect to the parabola `y^(2)=4ax` isA. a circleB. a parabolaC. a straight lineD. an ellipse |
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Answer» Correct Answer - B The equation of any tangent to `y^(2)=4ax` is `y=mx+a/m" ….(i)"` Let (h, k) be the ploe of (i) with respect to the parabola `y^(2)=4ax`. `ky=2b(x+y)` `"or, "2bx-ky+2ky=0" ....(ii)"` Clearly, (i) and (ii) represent the same straight line. `:." "(2b)/m=(-k)/(-1)=(2bh)/("a/m")` `rArr" "k=(2b)/m" and "m^(2)=a/h` `rArr" "m=(2b)/k" and "m^(2)=a/h` `rArr" "((2b)/k)^(2)=a/hrArrk^(2)=(4b^(2)h)/a` Hence, the locus of (h, k) is `y^(2)=(4b^(2)x)/a`, which is a parabola. |
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| 179. |
Given : A circle, `2x^2+""2y^2=""5`and a parabola, `y^2=""4sqrt(5)""x`.Statement - I : An equation of a common tangent to thesecurves is `y="x+"sqrt(5)`Statement - II : If the line, `y=m x+(sqrt(5))/m(m!=0)`is theircommon tangent, then m satisfies `m^4-3m^2+""2""=0.`(1)Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I(2)Statement -I is True; Statement -II is False.(3)Statement -I is False; Statement -II is True(4)Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-IA. Statement-1 is True, Statement - 2 is true, Statement-2 is a correct explanation for Statement-1`B. Statement-1 is True, Statement - 2 is true, Statement-2 is not a correct explanation for Statement-7C. Statement-1 is True, Statement - 2 is False.D. Statement-1 is True, Statement - 2 is True. |
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Answer» Correct Answer - B The equation of a tangent to `y^(2)=4sqrt5x` is `y=mx+sqrt5.m` where m is is the slope of the tangent. If it touches the circle `2x^(2)+2y^(2)=5`, then `|(sqrt(5)//m)/(sqrt(1+m^(2)))|=sqrt((5)/(2))` ` implies m sqrt(1+m^(2))=sqrt(2)` `implies m^(4)+m^(2)-2=0` `implies (m^(2)+2)(m^(2)-1)=0` `implies m = +- 1` Substituting these values in `y=x+sqrt5" and "y=-xsqrt5`. Also, values of m satify `m^(4)-3m^(3)+2=0.` Hence, both the statements are true. But, statement II is not a correct explanation of statement-i. |
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| 180. |
Statement-1: `y+b=m_(1) (x+a)` and `y+b=m_(2)(x+a)` are perpendicular tangents to the parabola `y^(2)=4ax`. Statement-2: The locus of the point of intersection of perpendicular tangents to a parabola is its directrix.A. Statement-1 is True, Statement - 2 is true, Statement-2 is a correct explanation for Statement-1`B. Statement-1 is True, Statement - 2 is true, Statement-2 is not a correct explanation for Statement-6C. Statement-1 is True, Statement - 2 is False.D. Statement-1 is True, Statement - 2 is True. |
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Answer» Correct Answer - A Statement-2 is true. Statement-1 is also true, because given tangents are drawn from the point (-a, -b) which is a point on the directrix. Also, statement-2 is a correct explanation for statement-1. |
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| 181. |
A circle drawn on any focal AB of the parabola `y^(2)=4ax` as diameter cute the parabola again at C and D. If the parameters of the points A, B, C, D be `t_(1), t_(2), t_(3)" and "t_(4)` respectively, then the value of `t_(3),t_(4)`, isA. -1B. 2C. 3D. none of these |
| Answer» Correct Answer - C | |
| 182. |
Let P be thepoint on the parabola, `y^2=8x`which is at a minimum distancefrom the centre C of thecircle,`x^2+(y+6)^2=1.`Then the equation of the circle, passing through C and having its centre at P is :(1) `x^2+y^2-4x+8y+12=0`(2) `x^2+y^2-x+4y-12=0`(3) `x^2+y^2-x/4+2y-24=0`(4) `x^2+y^2-4x+9y+18=0`A. `x^(2)+y^(2)-x+4y-12=0`B. `x^(2)+y^(2)=(1)/(4)x+2y-24=0`C. `x^(2)+y^(2)-4x+9y+12=0`D. `x^(2)+y^(2)-4x+8y+12=0` |
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Answer» Correct Answer - D Let `P(2t^(2), 4t)` be a point on `y^(2)=8x`. The coordinates of the centre C are(0, -6). `:." "CP^(2)=4t^(4)+(4t+6)^(2)` `"Let z =CP"^(2)` Then, `z=4t^(4)+4(2t+3)^(2)` `:." "(dz)/(dt)=16t^(3)+16(2t+3)" and "(d^(2)z)/(dt^(2))=48t^(2)+32` For maximum or minimum values of z, we must have `(dz)/(dt)=0rArr16(t^(3)+2t+3)=0` `rArr" "(t+1)(t^(2)-t+3)=0rArrt=-1` Clearly, `((d^(2)z)/(dt^(2)))_(t=-1)=80gt." So, z or CP is minimum when"` t=-1. So, the coordinates of P are (2, -4). The equation of the circle having centre at P(2, -4) adn passing through C(0, -6) is `(x-2)^(2)+(y+4)^(2)=(0-2)^(2)+(-6+4)^(2)` `"or, "x^(2)+y^(2)-4x+8y+12=0` ALITER Substituting t=2 in (i), the equation of the normal at P(4, 2) is y + 2x=12 whose x intercep is 6. The tangent to the circle at Q is perpendicular to the normal `y+2x==12`, So, the slope of the tangent is `1/2`. |
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| 183. |
`P` and `Q` are two distinct points on the parabola, `y^2 = 4x` with parameters `t` and `t_1` respectively. If the normal at `P` passes through `Q`, then the minimum value of `t_1 ^2` isA. 4B. 6C. 8D. 2 |
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Answer» Correct Answer - C The coordinates of P and Q are `(t^(2), 2t)` and `(t_(1)^(2), 2t_(1))` respectively. The normal at P passes through Q. `:." "t_(1)=-t-2/trArrt_(1)^(2)rArrt_(1)^(2)+4/t^(2)+4` `"Now, "AM geGM` `rArr" "t^(2)+4/t^(2)ge2sqrt(t^(2)xx4/t^(2))rArrt^(2)+4/t^(2)ge4rArrt_(1)^(2)ge8` Hence, the minimum value of `t_(1)^(2)` is 8. |
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| 184. |
Area of the triangle formed by the vertex, focus and one end of latusrectum of the parabola `(x+2)^(2)=-12(y-1)` isA. 36B. 18C. 9D. 6 |
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Answer» Correct Answer - C (3) Consider equal parabola `y^(2)=12x`. Vertex: V(0,0) Focus: S(3,0) End point of Lutus Rectum : A(3,6) `:." Required area"=(1)/(2)SVxxAS=9` sq. units |
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| 185. |
The mid-point of the chord `2x+y-4=0` of the parabola `y^(2)=4x` isA. (5/2, -1)B. (-1, 5/2)C. (3/2, -1)D. none of these |
| Answer» Correct Answer - A | |
| 186. |
The conic represented by the equation `x^(2)+y^(2)-2xy+20x+10=0,` isA. Pair of straight linesB. CircleC. ParaabolaD. Ellipse |
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Answer» Correct Answer - C Comparing the given equation with the equation `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0,` we have `a=1, b=1, h=-1, c=10, g =10" and "f=0` `:." "abc+2fgh-af^(2)-bg^(2)-ch^(2)=10-100=-100ne0.` and, `h^(2)-ab=(-1)^(2)-1xx1=0rArrh^(2)=ab` Thus, we have `Deltane0" and "h^(2)=ab` So, the given equation represents a parabola. |
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| 187. |
The two ends of latusrectum of a parabola are the points (3, 6) and (-5, 6). The focus, isA. (1, 6)B. (-1, 6)C. (1, -6)D. (-1, -6) |
| Answer» Correct Answer - B | |
| 188. |
Find the vertex, the co-ordinate of focus, the equation of the directrix and the length of latus rectum (i) y^2=10x (ii) x^2=6y |
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Answer» (i) y2 = 10 x A. Compare this equation with ( y - k )2 = 4a( x - h ) then we get k = 0 , h = 0 . So vertex = ( h , k ) = ( 0 , 0 ) Also 4a = 10 ( as per given data ) . So a = 10 / 4 = 5 / 2 As we know that focus of the parabola is S( a, 0 ) [since a > 0]. Then coordinates of S = ( 5 / 2 , 0 ) Since the given parabola is symmetric about x axis and S lies on x axis Equation of directrix is x + a = 0 , i.e x + 5 / 2 = 0 , 2x + 5 = 0 we know that length of latus rectum is 4a . So length = 4( 5 / 2 ) = 10 (ii) x2 = 6y A. Compare the given equation with ( x - h )2 = 4a( y - k ) then we get h = 0 , k = 0 .So vertex = ( h , k ) = ( 0 , 0 ) Also 4a = 6 ( as per given data ) . So a = 6 / 4 = 3 / 2 As we know that focus of this parabola is ( 0 , a ) [ since a > 0 ] Then S ( 0 , 3 / 2 ) Since the given parabola is symmetric about y axis and S lies on y axis Equation of directrix = y + a = 0 , i.e y + 3 / 2 = 0 , 2y + 3 = 0 we know that length of the latus rectum = 4a . So length = 4( 3 / 2 ) = 6 |
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| 189. |
If the parabola `y^(2)=4ax` passes through (3, 2). Then the length of its latusrectum, isA. `2/3`B. `4/3`C. `1/3`D. `4` |
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Answer» Correct Answer - B It is given the parabola `y^(2)=4 ax` passes through (3, 2). `:." "4=12arArra=1//3` Hence, length of L.R.=4 `a = 4/3` |
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| 190. |
The normals at the ends of the latusrectum of the parabola `y^(2)=4ax" are (a, 2a) and (a, -2a)"`.A. `pi/6`B. `pi/4`C. `pi/3`D. `pi/2` |
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Answer» Correct Answer - D The coordinates of the end points of the latusrectum of the parabola `y^(2)=4ax` are (a, 2a) and (a, -2a). The equation of the normal at (a, 2a) is `y-2a=-(2a)/(2a)(x-a)or, x+y-3a=0" ...(i)"` The equation of the normal at (a, -2a) is `y+2a=(-2a)/(2a)(x-a)or, x-y-3a=0" ...(ii)"` Clearly, (i) and (ii) are perpendicular as the product of their slopes is -1. |
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| 191. |
If l denotes the semi-latusrectum of the parabola`y^(2)=4ax,` and SP and SQ denote the segments of and focal chord PQ, being the focusm the SP, I, SQ are in the relationA. A.P.B. G.P.C. H.P.D. `t^(2)=SP^(2)+SQ^(2)` |
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Answer» Correct Answer - C From the above property I is the HM so SP and SQ. |
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| 192. |
If a focal chord of `y^2=4a x`makes an angle `alpha in [0,pi/4]`with the positive direction of the x-axis, then find the minimum length ofthis focal chord.A. `4a sec^(2)alpha`B. `2a "cosec"^(2)alpha`C. `4a" cosec"^(2)alpha`D. `4a cot^(2)alpha` |
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Answer» Correct Answer - C Let `P(at_(1)^(2), 2at_(1))" and Q"(at_(2)^(2), 2at_(2))` be the end points of a focal chord PQ which makes an angle `alpha` with the axis of the parabola. Then, `PQ=a(t_(2)-t_(1))^(2)` Also, `tan alpha" = slope of PQ"` `rArr" "tan alpha=2/(t_(2)+t_(1))rArrt_(1)+t_(2)=2 cot alpha` `:." "PQ=a(t_(2)-t_(1))^(2)` `rArr" "PQ=a{(t_(2)+t_(1))^(2)-4t_(1)t_(2)}` `rArr" "PQ=a {4 cot^(2)alpha+4}" "[becauset_(2)+t_(1)=2 cot alphaandt_(1)t_(2)=-1}` `rArr" "PQ=4a" cosec"^(2)alpha.` |
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| 193. |
If 3x+4y+k=0 represents the equation of tangent at the vertex of the parabola `16x^(2)-24xy^(2)+14x+2y+7=0`, then the value of k is ________ . |
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Answer» Correct Answer - 3 3 Given that equation of tangent at vertex is 3x+4y+k=0. Let the equation of axis be `4x-3y+lamda=0` Therefore, equation of parabola is `(4x-3y+lamda)^(2)=mu(3x+4y+k)` (1) Given equation of parabola is `16x^(2)-24xy+9y^(2)+14x+2y+7=0` (2) Comparing equations (1) and (2), we get `8lamda-3mu=14,-6lamda-4mu=2,lamda^(2)-kmu=7` Solving, we get `lamda=1,mu=-2andk=3`. |
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| 194. |
If the length of a focal chord of the parabola `y^2=4a x`at `a`distance `b`from the vertex is `c ,`then prove that `b^2c=4a^3dot`A. `2a^(2)=bc`B. `a^(3)=b^(2)c`C. `ac=b^(2)`D. `b^(2)c=4a^(3)` |
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Answer» Correct Answer - D Let `P(at_(1)^(2), 2at_(1))" and "Q(at_(2)^(2), 2at_(2))` be the end points of a focal chord of the parabola `y^(2)=4ax`. Then, `PQ=a(t_(2)-t_(1))^(2)` The equation of PQ is `(t_(1)+t_(2))y=2x-2a" "[becauset_(1)ty_(2)=-1]` It is given that PQ=c, and it is at distance b form the vertex. `:." "a(t_(2)-t_(1))^(2)n =c" and, "|(-2a)/(sqrt((t_(1)-t_(2))^(2)+4))|=b` `rArr" "(t_(1)-t_(2))^(2)=c/a" and, "(2a)/((t_(1)-t_(2)))=b" "[becauset_(1)t_(2)=-1]` `rArr" "((2a)/a)^(2)=c/arArr4a^(3)=b^(2)c` |
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| 195. |
Length of the focal chord of the parabola `(y +3)^(2) = -8(x-1)` which lies at a distance 2 units from the vertex of the parabola isA. 8B. `6sqrt(2)`C. 9D. `5sqrt(3)` |
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Answer» Correct Answer - A Lengths are invariant under change of axes. Consider `y^(2) = 8x`. Focus is (2,0). So, focal chord at distance "2" than vertex is latus rectum which has length "8". |
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| 196. |
The focus of the parabola `y^(2)-x-2y+2=0` isA. `(1//4,0)`B. `(1//2)`C. `(3//4,1)`D. `(5//4, 1)` |
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Answer» Correct Answer - D We have, `y^(2)-2y=x-2rArr(y-1)^(2)=x-1` The coordinates of the focus ot this parabola are given by `x-1=1/4,y-1=0" "[because" "y^(2)=4ax" has focus at (a, 0)"]` `rArr" "x=5//4, y=1` Hence, the coordinates of the focus are (5/4, 1). |
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| 197. |
The directrix of the parabola `x^2-4x-8y + 12=0` isA. y=0B. x=1C. y= -1D. x = -1 |
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Answer» Correct Answer - C We have, `x^(2)-4x-8y+12=0rArr(x-2)^(2)=8(y-1)" …(i)"` Clearly, it represents a parabola with vertex at (2, 1) and latusrectum = 8. Shifting the origin at (2, 1), we have `x=X+2" and "y=Y+1" ….(ii)"` Substituting these in (i), we get `X^(2)=8Y` It represents a parabola whose directrix has the equation `Y=-2" "[because" y=-a is the directrix "x^(2)="4ny"]` `"or, "y=-1" "["Using (ii)"]` |
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| 198. |
The directrix of the parabola `x^2-4x-8y + 12=0` isA. 4B. 6C. 8D. 10 |
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Answer» Correct Answer - C The equation of the parabola is `x^(2)-4x-8y+12=0` `rArr" "(x-2)^(2)=8y-8rArr(x-2)^(2)=8(y-1)` Clearly, it represents a parabola having vertex at (2, 1) and length of the latusretum = 8. |
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| 199. |
If (0, 3) and (0, 2) are respectively the vertex and focus of a parabola, then its equation, isA. `x^(2)+8y=12`B. `y^(2)+8x=32`C. `x^(2)-8y=32`D. `y^(2)-8x=32` |
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Answer» Correct Answer - A Clearly, vertex (0, 4) and focus (0, 2) lie on y-axis So, axis of the parabola is along y-axis and it opens downward. The distance between vertex and focus is 2 units. `:." Latusrectum = 4 "xx"2 = 8"` Hence, equation of the parabola is `(x-0)^(2)=-8(y-4)" or , "x^(2)+8y=32` |
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| 200. |
The axis of the parabola `9y^2-16x-12y-57 = 0` isA. `3y=2`B. `x+3y=3`C. `2x=3`D. `y=3` |
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Answer» Correct Answer - A We have `9y^(2)-16x-12y-57=0` `rArr" "y^(2)-4/3y=16/9x+57/9` `rArr" "(y-2/3)^(2)=(16x)/9+61/9rArr(y-2/3)^(2)=16/9(x+61/16)` The axis of this parabola is `y=2/3=0rArr3y=2" "|because" Axis of "y^(2)="4ax is y=0" |` |
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