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JEE • MATHEMATICS • PARABOLA

Parabola practice · Medium

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Showing 121-150 of 211 questions

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121 Medium
121 of 211
Application +4 | -1
The focus of the parabola x2=2x+2yx^2=2 x+2 y is
View answer and explanation
Correct option: C
(1,0)(1,0)
Option B ((1, −12\frac{{-1}}{{2}})) is incorrect. Option C (1,0) is correct because by completing the square, the equation can be rewritten in standard form, revealing the vertex and thus the focus.
122 Medium
122 of 211
Application +4 | -1
The length of the latus rectum of the parabola x2−4x−8yx^2-4 x-8 y +12=0+12=0 is
View answer and explanation
Correct option: C
8
Option C (8) is correct because the length of the latus rectum is found by rewriting the equation in standard form and identifying the coefficient of the linear term.
123 Medium
123 of 211
Application +4 | -1
At what point on the parabola y2=4xy^2=4 x, the normal makes equal angles with the coordinate axes
View answer and explanation
Correct option: D
(1,−2)(1,-2)
Option D (x2+y2−30x+12y−27=0x^2+y^2-30 x+12 y-27=0) is correct because it's the equation of the circle with the normal chord as diameter, derived from the given points.
124 Medium
124 of 211
Application +4 | -1
PQP Q is a double-ordinate of the parabola y2=4axy^2=4 a x. The locus of the points of trisection of PQP Q is
View answer and explanation
Correct option: A
9y2=4ax9 y^2=4 a x
Option A (9y2=4ax9 y^2=4 a x) is correct because the locus of the points of trisection of a double ordinate of the parabola y2=4axy^2 = 4ax is found by scaling the y-coordinate.
125 Medium
125 of 211
Application +4 | -1
If the line lx+my+n=0l x+m y+n=0 is a tangent to the parabola y2=4axy^2=4 a x, then the locus of its point of contact is
View answer and explanation
Correct option: C
A parabola
Option C (A parabola) is correct because eliminating 't' from the tangent equation results in a parabolic relationship between m, n, and l.
126 Medium
126 of 211
Application +4 | -1
The vertex of the parabola y2=8xy^2=8 x is at the centre of a circle and the parabola cuts the circle at the ends of its latusrectum. Then, the equation of the circle is
View answer and explanation
Correct option: B
x2+y2=20x^2+y^2=20
Option B (x2+y2=20x^2+y^2=20) is correct because the circle passes through the ends of the latus rectum, and its center is at the vertex of the parabola.
127 Medium
127 of 211
Application +4 | -1
The equation of the tangent to the parabola y2=16xy^2=16 x, which is perpendicular to the line y=3x+7y=3 x+7 is
View answer and explanation
Correct option: D
3y+x+36=03 y+x+36=0
Option D (3y+x+36=03y+x+36=0) is correct because it is the equation of the tangent to the parabola y2=16xy^2=16x that is perpendicular to the line y=3x+7y=3x+7.
128 Medium
128 of 211
Application +4 | -1
The straight line y=m(x−a)y=m(x-a) meets the parabola y2=4axy^2=4 a x in two distinct points for
View answer and explanation
Correct option: C
All mI^R−{0}{m} \hat{{I}} {R}-\{0\}
Option C (All mI^R−{0}{m} \hat{{I}} {R}-\{{0}\}) is correct because the line intersects the parabola at two distinct points for all non-zero values of m.
129 Medium
129 of 211
Application +4 | -1
The equation of diameter of parabola y2=xy^2=x corresponding to the chord x−y+1=0x-y+1=0 is
View answer and explanation
Correct option: B
2y=12 y=1
Option B (2y=12y=1) is correct because the equation of the diameter of the parabola is found using the formula y=2amy = \frac{{2a}}{{m}}.
130 Medium
130 of 211
Application +4 | -1
The co-ordinate of the vertex of parabola y=x2+bx+cy=x^2+b x+c is (1,2)(1,2) then find value of bb.
View answer and explanation
Correct option: B
−2-2
Option B (-2) is correct because the x-coordinate of the vertex of the parabola y=x2+bx+cy = x^2 + bx + c is given by −b/2-b/2, and since the vertex is (1, 2), we have −b/2=1-b/2 = 1, so b=−2b = -2.
131 Medium
131 of 211
Application +4 | -1
The equation of the parabola whose vertex and focus lies on the xx-axis at distance aa and a′a^{\prime} from the origin is
View answer and explanation
Correct option: A
y2=4(a′−a)(x−a)y^2=4\left(a^{\prime}-a\right)(x-a)
Option A (y2=4(a′−a)(x−a)y^2=4(a^{\prime}-a)(x-a)) is correct because it represents a parabola with vertex at (a, 0) and focus at (a', 0).
132 Medium
132 of 211
Application +4 | -1
The focus of the parabola y2−x−2y+2=0y^2-x-2 y+2=0 is
View answer and explanation
Correct option: D
(5/4,1)(5 / 4,1)
Option D (5/4, 1) is correct because the focus of the parabola is found by rewriting the equation in standard form and applying the formula for the focus.
133 Medium
133 of 211
Application +4 | -1
A ray of light moving parallel to the xx-axis gets reflected from a parabolic mirror whose equation is y2=4(x−1)y^2=4(x-1). After reflection the ray passes through
View answer and explanation
Correct option: A
(2,0)(2,0)
Option A ((2,0)) is correct because a ray parallel to the x-axis, after reflection from a parabola, passes through its focus, which is (2,0) in this case.
134 Medium
134 of 211
Application +4 | -1
The angle of intersection between the curves y2=4xy^2=4 x and x2=32yx^2=32 y at point (16,8)(16,8), is
View answer and explanation
Correct option: A
tan⁡−1(35)\tan ^{-1}\left(\frac{3}{5}\right)
Option A (tan⁡−1(35)\tan ^{-1}\left(\frac{3}{5}\right)) is correct because it represents the angle of intersection between the curves y2=4xy^2=4 x and x2=32yx^2=32 y at the point (16,8).
135 Medium
135 of 211
Application +4 | -1
If the normals at two points PP and QQ of a parabola y2=4ax{y^2} = 4ax intersect at a third point RR on the curve, then the product of ordinates of PP and QQ is
View answer and explanation
Correct option: D
8a28{a^2}
Option D (8a28{a^2}) is correct because the product of the ordinates of P and Q is calculated using the properties of normals to a parabola.

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136 Medium
136 of 211
Application +4 | -1
The slope of tangents drawn form a point (4,  10)\left( {4,\;10} \right) to the parabola y2=9x{y^2} = 9x are
View answer and explanation
Correct option: B
14,94\frac{1}{4},\frac{9}{4}
Option B (14,94\frac{1}{4},\frac{9}{4}) is correct because these are the slopes of the tangent lines to the parabola that pass through the given point (4, 10).
137 Medium
137 of 211
Analyzing +4 | -1
Let PP be the point on the parabola y2=8xy^2=8 x which is at a minimum distance from the centre C of the circle x2+(y+6)2=1x^2+(y+6)^2=1. Then the equation of the circle, passing through CC and having centre at PP is:
View answer and explanation
Correct option: D
x2+y2−4x+8y+12=0x^2+y^2-4 x+8 y+12=0
Option D (x2+y2−4x+8y+12=0x^2+y^2-4 x+8 y+12=0) is correct because it represents the equation of the circle passing through C and having its center at the point P on the parabola that is closest to C.
138 Medium
138 of 211
Application +4 | -1
Common tangent is drawn to y2=16xy^2=16 x and x2+y2=8x^2+y^2=8. Find square of distance between point of contact of common tangent on both the curves.
View answer and explanation
Correct option: B
72
Option B (72) is correct because it is the square of the distance between the points of contact of the common tangent on the parabola and the circle, calculated using the properties of tangents and circles.
139 Medium
139 of 211
Analyzing +4 | -1
Two parabolas have the same focus. If their directrices are the XX-axis and the YY-axis, respectively, then the slope of their common chord is
View answer and explanation
Correct option: A
±1\pm 1
Option C (3) is correct because solving for the intersection points and using the tangent line equation leads to a cubic equation with a = 3 as a valid solution.
140 Medium
140 of 211
Application +4 | -1
Equation of the common tangent to the circle x2+y2−6x=0x^2+y^2-6 x=0 and the parabola y2=4xy^2=4 x, 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 it is the tangent to both the circle and the parabola, found by equating the conditions for tangency.
141 Medium
141 of 211
Application +4 | -1
Locus of middle point of the chord of the parabola y2=4xy^2=4 x which passes through a point (1,2)(1,2)
View answer and explanation
Correct option: B
(y−1)2=(2x−1)(y-1)^2=(2 x-1)
Option B ((y−1)2=(2x−1)(y-1)^2=(2 x-1)) is correct because the locus of the midpoint of the chord is found using T=S1 and the condition that the chord passes through (1,2).
142 Medium
142 of 211
Application +4 | -1
The focus of the parabola y=2x2+xy=2 x^2+x is
View answer and explanation
Correct option: C
(−14,0)\left(-\frac{1}{4}, 0\right)
Option D (-1/4, 0) is correct because the focus of the parabola is found by rewriting the equation in standard form and applying the formula for the focus.
143 Medium
143 of 211
Remembering +4 | -1
The vertex of the parabola (y−2)2=16(x−1)(y-2)^2=16(x-1) is
View answer and explanation
Correct option: D
(1,2)(1,2)
Option D ((1,2)) is correct because the vertex of the parabola (y−k)2=4a(x−h)(y-k)^2=4a(x-h) is (h, k), and in this case, h=1 and k=2.
144 Medium
144 of 211
Application +4 | -1
If the parabola x2=ayx^2= ay makes an intercept of length 40\sqrt{40} on the line y−2x=1y-2 x=1, then a=a=
View answer and explanation
Correct option: B
−2-2
Option B (-2) is correct because solving the quadratic equation derived from the intersection of the parabola and line yields a=-2 as a valid solution.
145 Medium
145 of 211
Application +4 | -1
The length of the latusrectum of the parabola 169{(x−1)2+(y−3)2}=(5x−12y+17)2169\left\{ {{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}} \right\} = {\left( {5x - 12y + 17} \right)^2} is
View answer and explanation
Correct option: C
28/1328/13
Option C (28/1328/13) is correct because the length of the latusrectum of the parabola is twice the distance of the focus from the directrix.
146 Medium
146 of 211
Application +4 | -1
The area of the triangle formed by tangents and the chord of contact from (3,4)(3,4) to y2=2xy^2=2 x is
View answer and explanation
Correct option: A
101010 \sqrt{10}
Option A (101010 \sqrt{10}) is correct because the area of the triangle formed by tangents and the chord of contact is calculated using the formula (S11)322a\frac{{(S_{11})^{\frac{{3}}{{2}}}}}{{2a}}.
147 Medium
147 of 211
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.
148 Medium
148 of 211
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.
149 Medium
149 of 211
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.
150 Medium
150 of 211
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.
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