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JEE • MATHEMATICS • PARABOLA

Parabola practice · Easy

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91 Easy
91 of 356
Application +4 | -1
The slope of the normal at the point (at2,2at)\left( {a{t^2},2at} \right) of the parabola, y2=4ax,{y^2} = 4ax, is
View answer and explanation
Correct option: C
−t- t
Option C (-t) is correct because the slope of the normal is the negative reciprocal of the slope of the tangent, which is -t.
92 Easy
92 of 356
Application +4 | -1
If the normal at (ap2,2ap)\left( {a{p^2},2ap} \right) on the parabola y2=4ax{y^2} = 4ax, meets the parabola again at (aq2,2aq),\left( {a{q^2},2aq} \right), then
View answer and explanation
Correct option: A
p2+pq+2=0{p^2} + pq + 2 = 0
Option A (p2+pq+2=0{p^2} + pq + 2 = 0) is correct because it is derived from the condition for the normal at a point on the parabola to meet the parabola again.
93 Easy
93 of 356
Understanding +4 | -1
The maximum number of normal that can be drawn from a point to a parabola is
View answer and explanation
Correct option: D
3
Option D (3) is correct because a cubic equation can have at most three real roots, and the equation for the normal to a parabola is cubic.
94 Easy
94 of 356
Analyzing +4 | -1
If the tangents and normals at the extremities of a focal chord of a parabola intersect at (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) respectively, then
View answer and explanation
Correct option: C
y1=y2y_1=y_2
Option C (y1=y2y_1=y_2) is correct because the y-coordinates of the intersection points of tangents and normals at the extremities of a focal chord are equal.
95 Easy
95 of 356
Analyzing +4 | -1
If the axis of the parabola y2=4axy^2=4 a x, its directrix and the chord joining t1t_1 and t2t_2 are concurrent, then t1t2=t_1 t_2=
View answer and explanation
Correct option: B
1
Option A (1 / 2,1 / 2) is incorrect. The correct answer is C (-1) because it is derived from the condition that the axis, directrix, and chord joining the points are concurrent.
96 Easy
96 of 356
Analyzing +4 | -1
The curve described parametrically by x=t2+2t−1,  y=3t+5x = {t^2} + 2t - 1,\;y = 3t + 5 represents
View answer and explanation
Correct option: C
A parabola
Option C (A parabola) is correct because eliminating the parameter 't' results in an equation of a parabola.
97 Easy
97 of 356
Application +4 | -1
The latusrectum of a parabola whose focal chord is PSQP S Q such that SP=3S P=3 and SQ=2S Q=2, is given by
View answer and explanation
Correct option: A
24/524 / 5
Option A (24/524 / 5) is correct because the latus rectum is twice the semi-latus rectum, which is the harmonic mean of the segments of the focal chord.
98 Easy
98 of 356
Application +4 | -1
The equation of the latusrectum of the parabola x2+4x+2y=0{x^2} + 4x + 2y = 0, is equal to
View answer and explanation
Correct option: C
2y=32y = 3
Option C (2y=32y = 3) is correct because it represents the equation of the latus rectum after completing the square and shifting the parabola to its standard form.
99 Easy
99 of 356
Application +4 | -1
The vertex of the parabola x2+8x+12y+4=0x^2+8 x+12 y+4=0 is
View answer and explanation
Correct option: A
(−4,1)(-4,1)
Option A ((-4, 1)) is correct because the vertex of the parabola is (-4, 1) after rewriting the equation in standard form.
100 Easy
100 of 356
Application +4 | -1
On the parabola y=x2y = {x^2}, the point at least distance from the straight line y=2x−4y = 2x - 4 is
View answer and explanation
Correct option: A
(1,1)\left( {1,1} \right)
Option A ((1,1)) is correct because it represents the point on the parabola where the tangent is parallel to the given line, minimizing the distance.
101 Easy
101 of 356
Application +4 | -1
The tangents drawn from the ends of latusrectum of y2=12x{y^2} = 12x meets at
View answer and explanation
Correct option: A
Directrix
Option A (Directrix) is correct because the tangents at the ends of the latus rectum of a parabola intersect at the directrix.
102 Easy
102 of 356
Application +4 | -1
The intersection point of the normals drawn at the end points of latusrectum of the parabola x2=−2y{x^2} = - 2y is
View answer and explanation
Correct option: D
(0,−32)\left( {0, - \frac{3}{2}} \right)
Option D (0,−32)\left( {0, - \frac{{3}}{{2}}} \right) is correct because it is the intersection point obtained by solving the equations of the normals at the end points of the latus rectum.
103 Easy
103 of 356
Analyzing +4 | -1
Let the line lx+my=1lx+m y=1 cut the parabola y2=4axy^2=4 a x in the points AA and BB. Normals at AA and BB meet at point CC. Normal from CC other than these two meet the parabola at DD, then the coordinate of DD are
View answer and explanation
Correct option: D
(4am2l2,4aml)\left(\frac{4 a m^2}{l^2}, \frac{4 a m}{l}\right)
Option B (3−223-2 \sqrt{2}) is correct because it represents the minimum squared distance between the circle and the parabola, calculated using the common normal.
104 Easy
104 of 356
Application +4 | -1
If vertex of a parabola is (2,−1)(2,-1) and the equation of its directrix is 4x−3y=214 x-3 y=21, then the length of its latus rectum is:
View answer and explanation
Correct option: B
8
Option B (8) is correct because the length of the latus rectum is four times the distance from the vertex to the directrix.
105 Easy
105 of 356
Application +4 | -1
A tangent to the parabola y2=8xy^2=8 x makes an angle of 45∘45^{\circ} with the straight line y=3x+5y=3 x+5, then the equation of tangent is
View answer and explanation
Correct option: C
2x+y+1=02 x+y+1=0
Option C (2x+y+1=02 x+y+1=0) is correct because it represents the equation of the tangent to the parabola y2=8xy^2=8 x that makes an angle of 45∘45^{\circ} with the straight line y=3x+5y=3 x+5.

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106 Easy
106 of 356
Analyzing +4 | -1
The equation (x−2)2+(y−3)2=(3x+4y−25)2{\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = {\left( {\frac{{3x + 4y - 2}}{5}} \right)^2} represents
View answer and explanation
Correct option: A
A parabola
Option A (A parabola) is correct because the equation represents the set of points equidistant from a point (2,3) and a line 3x + 4y - 2 = 0, which is the definition of a parabola.
107 Easy
107 of 356
Application +4 | -1
Equation of the line joining the foci of the parabolas y2=4xy^2=4 x and x2=−4yx^2=-4 y is
View answer and explanation
Correct option: B
x−y−1=0x-y-1=0
Option B (x−y−1=0x-y-1=0) is correct because it represents the line joining the foci of the given parabolas, calculated by finding the coordinates of the foci and then the equation of the line.
108 Easy
108 of 356
Application +4 | -1
For the parabola y2=−12xy^2=-12 x, equation of directrix is x=ax=a. The value of 'aa' is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the value of 'a' in the directrix equation is 3, derived from the given parabola equation.
109 Easy
109 of 356
Application +4 | -1
Let PP be a point on the parabola x2=4yx^2=4 y. If the line segment joining A(0,−1)A(0,-1) and PP is divided by point QQ in the ratio 1:21: 2, then the locus of point QQ, is
View answer and explanation
Correct option: B
9x2=12y+89 x^2=12 y+8
Option B (9x2=12y+89 x^2=12 y+8) is correct because it represents the locus of point Q derived from the given conditions of the parabola and ratio division.
110 Easy
110 of 356
Analyzing +4 | -1
The tangent PTPT and the normal PNPN to the parabola y2=4axy^2=4 a x at a point PP on it meet its axis at points TT and NN, respectively. The locus of the centroid of ΔPTN\Delta PTN is a parabola whose
View answer and explanation
Correct option: A
Vertex is (2a/3,0)(2 a / 3,0)
Option A (Vertex is (2a/3,0)(2 a / 3,0)) is correct because the locus of the centroid is a parabola with vertex at (2a/3,0)(2a/3, 0).
111 Easy
111 of 356
Application +4 | -1
The coordinates of a point on the parabola y2=8  x{y^2} = 8\;x whose focal distance is 4, are
View answer and explanation
Correct option: C
(2,±4)\left( {2, \pm 4} \right)
Option C ((2,±4)\left( {2, \pm 4} \right)) is correct because it's the point on the parabola with the given focal distance, calculated using the definition of focal distance.
112 Easy
112 of 356
Application +4 | -1
The directrix of the parabola 25[x2+y2−2y+1]=(3x+4y+1)225\left[x^2+y^2-2 y+1\right]=(3 x+4 y+1)^2 is
View answer and explanation
Correct option: A
3x+4y+1=03 x+4 y+1=0
Option A (3x+4y+1=03x+4y+1=0) is correct because it represents the directrix derived from the given parabola equation by comparing it to the standard form.
113 Easy
113 of 356
Application +4 | -1
The equation of a parabola having focus (3,0)(3,0) and directrix x+3=0x+3=0, is
View answer and explanation
Correct option: A
y2=12xy^2=12 x
Option A (y2=12xy^2=12 x) is correct because it satisfies the definition of a parabola with focus (3,0) and directrix x+3=0.
114 Easy
114 of 356
Analyzing +4 | -1
The number of real tangents that can be drawn to the curve y2+2xy+x2+2x+3y+1=0{y^2} + 2xy + {x^2} + 2x + 3y + 1 = 0 from the point (1,−2)\left( {1, - 2} \right) is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the point (1, -2) lies inside the parabola, and no real tangents can be drawn from an interior point.
115 Easy
115 of 356
Application +4 | -1
Two tangents are drawn from the point (−2,−1)\left( { - 2, - 1} \right) to the parabola y2=4x{y^2} = 4x. If α\alpha is the angle between these tangents, then tan⁡α\tan \mathrm{\alpha } is equal to
View answer and explanation
Correct option: A
3
Option A (3) is correct because solving for the slopes of the tangents and then applying the tangent formula yields tan α = 3.
116 Easy
116 of 356
Application +4 | -1
Equation of the parabola with its vertex at (1,1)\left( {1,1} \right) and focus (3,1)\left( {3,1} \right) is
View answer and explanation
Correct option: C
(y−1)2=8(x−1){\left( {y - 1} \right)^2} = 8\left( {x - 1} \right)
Option C ((y−1)2=8(x−1){(y - 1)^2} = 8(x - 1)) is correct because it represents a parabola with the given vertex and focus, opening to the right.
117 Easy
117 of 356
Application +4 | -1
A parabola has the origin as its focus and the line x=2x = 2 as the directrix. Then, the vertex of the parabola is at
View answer and explanation
Correct option: C
(1,0)\left( {1,0} \right)
Option C ((1,0)\left( {1,0} \right)) is correct because the vertex of a parabola is the midpoint between the focus and the directrix.
118 Easy
118 of 356
Application +4 | -1
The radical centre of the circles drawn on the focal chords of y2=4axy^2=4 a x as diameters, is
View answer and explanation
Correct option: C
(0,0)(0,0)
Option C ((0,0)) is correct because the radical center of circles drawn on focal chords as diameters is always at the origin.
119 Easy
119 of 356
Application +4 | -1
Equation of the parabola with its vertex at (1,1)\left( {1,1} \right) and focus (3,1)\left( {3,1} \right) is
View answer and explanation
Correct option: C
(y−1)2=8(x−1){\left( {y - 1} \right)^2} = 8\left( {x - 1} \right)
Option C ((y−1)2=8(x−1){{\left( {y - 1} \right)}^2} = 8\left( {x - 1} \right)) is correct because it represents the equation of a parabola with the given vertex and focus.
120 Easy
120 of 356
Application +4 | -1
If a focal chord of the parabola y2=ax{y^2} = ax is 2x−y−8=02x - y - 8 = 0, then the equation of the directrix is
View answer and explanation
Correct option: A
x+4=0x + 4 = 0
Option A (x+4=0x + 4 = 0) is correct because it is found by determining the value of 'a' from the focal chord and then using the formula for the directrix of the parabola.
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