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JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

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Showing 391-420 of 586 questions

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391 Medium
391 of 586
Application +4 | -1
The equation of common tangent to the circle x2+y2=2x^2+y^2=2 and parabola y2=8xy^2=8 x is
View answer and explanation
Correct option: B
y=x+2y=x+2
Option B (y=x+2y=x+2) is correct because it is the common tangent to both the circle x2+y2=2x^2+y^2=2 and the parabola y2=8xy^2=8x.
392 Easy
392 of 586
Application +4 | -1
The line y=2x+cy=2 x+c is tangent to the parabola y2=4xy^2=4 x, then what is the value of cc ?
View answer and explanation
Correct option: B
1/21 / 2
Option B (1/21/2) is correct because it satisfies the tangency condition for a line to be tangent to a parabola.
393 Medium
393 of 586
Application +4 | -1
The number of common tangents to parabola y2=8xy^2=8 x and hyperbola 3x2−y2=33 x^2-y^2=3 is
View answer and explanation
Correct option: C
2
Option C (2) is correct because by solving for the common tangents of the parabola and hyperbola, we find that there are two such tangents.
394 Medium
394 of 586
Application +4 | -1
Let OO be the origin and AA be a point on the curve y2=4x.{y^2} = 4x. Then, the locus of the mid point of OAOA, is
View answer and explanation
Correct option: D
y2=2x{y^2} = 2x
Option D (y2=2x{y^2} = 2x) is correct because the locus of the midpoint of a line segment joining the origin and a point on the curve y2=4x{y^2} = 4x is y2=2x{y^2} = 2x.
395 Easy
395 of 586
Application +4 | -1
If the parabola y2=4axy^2=4 a x passes through (3,2)(3,2), then the length of its latusrectum, is
View answer and explanation
Correct option: B
4/34 / 3
Option B (4 / 3) is correct because substituting (3,2) into the equation allows us to solve for 'a', and then calculate the latus rectum 4a.
396 Medium
396 of 586
Application +4 | -1
If the line y−3x+3=0y-\sqrt{3} x+3=0 cuts the parabola y2=x+2y^2=x+2 at AA and BB, then PA⋅PBP A \cdot P B is equal to [where, P=(3,0)P=(\sqrt{3}, 0) ]
View answer and explanation
Correct option: A
4(3+2)3\frac{4(\sqrt{3}+2)}{3}
Option A (4(3+2)3\frac{{4(\sqrt{{3}}+2)}}{{3}}) is correct because it represents the value of PA * PB calculated using parametric equations.
397 Easy
397 of 586
Application +4 | -1
If 2x+y+k=02x + y + k = 0 is a normal to the parabola y2=−8x,{y^2} = - 8x, then the values of kk is
View answer and explanation
Correct option: D
24
The correct answer is D (24) because it represents the value of k for which the given line is normal to the parabola.
398 Easy
398 of 586
Application +4 | -1
From the point (−1,−6)\left( { - 1, - 6} \right) two tangents are drawn to the parabola y2=4x{y^2} = 4x. Then, the angle between the two tangents is
View answer and explanation
Correct option: D
900{90^0}
The correct answer is D (900{90^0}) because the product of the slopes of the two tangents is -1, indicating they are perpendicular.
399 Easy
399 of 586
Application +4 | -1
The latus rectum of the parabola y2=5x+4y+1y^2=5 x+4 y+1 is
View answer and explanation
Correct option: C
5
Option C (5) is correct because after completing the square, the equation is in the form (y−k)2=4p(x−h)(y-k)^2 = 4p(x-h), and the latus rectum is 4p4p.
400 Easy
400 of 586
Application +4 | -1
The vertex of the parabola 9x2−6x+36y+9=09 x^2-6 x+36 y+9=0 is
View answer and explanation
Correct option: A
(1/3,−2/9)(1 / 3,-2 / 9)
Option A ((1/3,−2/9)(1 / 3,-2 / 9)) is correct because completing the square and rewriting the equation in vertex form reveals the vertex coordinates.
401 Easy
401 of 586
Application +4 | -1
The length of the latusrectum of the parabola 2{(x−a)2+(y−a)2}=(x+y)22\left\{(x-a)^2+(y-a)^2\right\}=(x+y)^2, is
View answer and explanation
Correct option: B
22a2 \sqrt{2} a
Option B (22a2 \sqrt{2} a) is correct because it is calculated using the distance between the focus and directrix of the parabola.
402 Medium
402 of 586
Application +4 | -1
If lx+my+n=0l x+m y+n=0 is tangent to the parabola x2=yx^2=y, then the condition of tangency is
View answer and explanation
Correct option: D
l2=4mnl^2=4 \mathrm{mn}
Option D (l2=4mnl^2=4 \mathrm{mn}) is correct because this is the condition derived when the discriminant of the quadratic equation formed by the intersection of the line and parabola is set to zero.
403 Easy
403 of 586
Application +4 | -1
If (0,6)\left( {0,6} \right) and (0,3)\left( {0,3} \right) are respectively the vertex and focus of a parabola, then its equation is
View answer and explanation
Correct option: A
x2+12y=72{x^2} + 12y = 72
Option A (x2+12y=72{x^2} + 12y = 72) is correct because it represents a parabola with the given vertex and focus.
404 Easy
404 of 586
Application +4 | -1
The normal to the parabola y2=8x{y^2} = 8x at the point (2,  4)\left( {2,\;4} \right) meets the parabola again at the point
View answer and explanation
Correct option: D
(18,  −12)\left( {18,\; - 12} \right)
Option D ((18,  −12)\left( {18,\; - 12} \right)) is correct because it's the point where the normal at (2, 4) intersects the parabola again, calculated using the parametric form of the parabola.
405 Medium
405 of 586
Application +4 | -1
The locus of the point of intersection of tangents to the parabola y2=4(x+1){y^2} = 4\left( {x + 1} \right) and y2=8(x+2){y^2} = 8\left( {x + 2} \right) which are perpendicular to each other is
View answer and explanation
Correct option: C
x+3=0x + 3 = 0
The correct answer is C (x+3=0x + 3 = 0) because the locus of the intersection of perpendicular tangents to parabolas is the directrix, and the directrix of the combined parabola system is x = -3.

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406 Easy
406 of 586
Application +4 | -1
The locus of the poles of tangents to the parabola y2=4axy^2=4 a x with respect to the parabola y2=4bxy^2=4 b x is
View answer and explanation
Correct option: B
a parabola
Option B (a parabola) is correct because the locus of the poles is found to be a parabola by relating the tangent equation to the polar equation.
407 Easy
407 of 586
Application +4 | -1
The equation of the parabola with focus (0,0)\left( {0,0} \right) and directrix x+y=4x + y = 4, is
View answer and explanation
Correct option: A
x2+y2−2xy+8x+8y−16=0{x^2} + {y^2} - 2xy + 8x + 8y - 16 = 0
Option A (x2+y2−2xy+8x+8y−16=0{x^2} + {y^2} - 2xy + 8x + 8y - 16 = 0) is correct because it is found by using the definition of a parabola: the distance to the focus equals the distance to the directrix.
408 Easy
408 of 586
Application +4 | -1
The point of contact of the tangent 18x−6y+1=018 x-6 y+1=0 to the parabola y2=2xy^2=2 x is
View answer and explanation
Correct option: D
(118,13)\left(\frac{1}{18}, \frac{1}{3}\right)
Option D (118,13)\left(\frac{{1}}{{18}}, \frac{{1}}{{3}}\right) is correct because it represents the point where the given tangent touches the parabola, found by comparing coefficients.
409 Easy
409 of 586
Application +4 | -1
The equation of the parabola with vertex at the origin and directrix y=2y = 2 is
View answer and explanation
Correct option: D
x2=−8y{x^2} = - 8y
Option D (x2=−8y{x^2} = - 8y) is correct because with the vertex at the origin and the directrix at y=2y = 2, the parabola opens downwards, and the equation is of the form x2=−4ay{x^2} = -4ay, where a=2a = 2.
410 Easy
410 of 586
Application +4 | -1
The tangents to a parabola at its vertex AA and any point PP meet at QQ. If SS be the focus of the parabola, then SP,SQS P, S Q, SAS A are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because SQ2=SP×SASQ^2 = SP \times SA, which is the condition for SP,SQ,SASP, SQ, SA to be in Geometric Progression.
411 Easy
411 of 586
Application +4 | -1
The points on the parabola y2=36xy^2=36 x, whose ordinate is three times the abscissa, are
View answer and explanation
Correct option: A
(0,0),(4,12)(0,0),(4,12)
Option A ((0,0),(4,12)) is correct because these points satisfy both the parabola's equation and the condition that the ordinate is three times the abscissa.
412 Medium
412 of 586
Application +4 | -1
Locus of mid point of chord of parabola x2=4yx^2=4 y having slope 2
View answer and explanation
Correct option: B
x=4x=4
Option B (x=4) is correct because the locus of the midpoint is found using T=S1 and the given slope condition.
413 Medium
413 of 586
Analyzing +4 | -1
The shortest distance between the line x−y=1 x\mathrm{ }-\mathrm{ }y\mathrm{ } = \mathrm{ }1\, and the curve x2=2y{x^2} = 2y is
View answer and explanation
Correct option: B
122\frac{1}{{2\sqrt 2 }}
Option B (122\frac{1}{{2\sqrt 2 }}) is correct because it is the shortest distance found by finding a tangent to the curve parallel to the line and then calculating the perpendicular distance.
414 Medium
414 of 586
Analyzing +4 | -1
The largest value of aa for which the circle x2+y2=a2x^2+y^2=a^2 falls totally in the interior of the parabola y2=4(x+4)y^2=4(x+4) is
View answer and explanation
Correct option: D
232 \sqrt{3}
Option D (232 \sqrt{{3}}) is correct because it represents the largest radius for the circle to be entirely within the parabola, found by setting the discriminant to zero.
415 Medium
415 of 586
Application +4 | -1
The length of the normal chord to the parabola y2=4xy^2=4 x, which subtends right angle at the vertex is
View answer and explanation
Correct option: A
636 \sqrt{3}
Option A (636 \sqrt{3}) is correct because it represents the length of the normal chord to the parabola that subtends a right angle at the vertex, calculated using the properties of the parabola and normal.
416 Easy
416 of 586
Application +4 | -1
If the normal chord of the parabola y2=4xy^2=4 x makes an angle of 45∘45^{\circ} with the axis of the parabola, then its length, is
View answer and explanation
Correct option: B
828 \sqrt{2}
Option B (828 \sqrt{2}) is correct because by finding the coordinates of the points where the normal chord intersects the parabola and calculating the distance between them.
417 Easy
417 of 586
Application +4 | -1
The equation of the common tangent to the circle (x−3)2+y2=9(x-3)^2+y^2=9 and the parabola y2=4xy^2=4 x above the xx-axis, is
View answer and explanation
Correct option: C
3y=x+3\sqrt{3} y=x+3
Option C (3y=x+3\sqrt{3} y=x+3) is correct because by finding the tangent to the parabola and then applying the condition for tangency to the circle, we arrive at this equation.
418 Easy
418 of 586
Application +4 | -1
The two parabolas x2=4y{x^2} = 4y and y2=4x{y^2} = 4x meet in two distinct points. One of these is the origin and the other is
View answer and explanation
Correct option: C
(4,4)\left( {4,4} \right)
Option C ((4,4)) is correct because it is the non-origin intersection point found by solving the system of equations for the two parabolas.
419 Medium
419 of 586
Application +4 | -1
If two tangents drawn from a point PP to the parabola y2=4x{y^2} = 4x be such that the slope of one tangent is double of the other, then PP lies on the curve
View answer and explanation
Correct option: B
9x=2y29x = 2{y^2}
Option D (None of these) is correct because the calculated common tangent does not match any of the provided options after correctly determining the value of m.
420 Easy
420 of 586
Application +4 | -1
The equation of parabola whose vertex and focus are (0,4)(0,4) and (0,2)(0,2), respectively, is
View answer and explanation
Correct option: C
x2+8y=32x^2+8 y=32
Option C (x2+8y=32x^2+8 y=32) is correct because it represents a parabola opening downwards with the given vertex and focus.
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