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JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

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Showing 301-330 of 586 questions

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301 Easy
301 of 586
Application +4 | -1
The point on parabola 2y=x22 y=x^2 which is nearest to the point (0,3)(0,3) is
View answer and explanation
Correct option: C
(±2,2)( \pm 2,2)
Option C ((±2,2)( \pm 2,2)) is correct because it minimizes the distance between the point and the parabola.
302 Medium
302 of 586
Analyzing +4 | -1
Two parabolas y2=4a(x−λ1)y^2=4 a\left(x-\lambda_1\right) and x2=4a(y−λ2)x^2=4 a\left(y-\lambda_2\right) always touch each other, λ1\lambda_1 and λ2\lambda_2 being variable parameters. Then, their points of contact lie on a
View answer and explanation
Correct option: D
hyperbola
Option D (hyperbola) is correct because the locus of the point of contact is found to be xy=4a2xy = 4a^2, which represents a hyperbola.
303 Easy
303 of 586
Application +4 | -1
If the point (a,2a)(a, 2 a) is an interior point of the region bounded by the parabola y2=16xy^2=16 x and the double ordinate through the focus, then
View answer and explanation
Correct option: B
a∈(0,4)a \in(0,4)
Option B (a∈(0,4)a \in (0,4)) is correct because the point must lie inside both the parabola and the region bounded by the double ordinate.
304 Easy
304 of 586
Application +4 | -1
The equation of the directrix of (x−1)2=2(y−2){\left( {x - 1} \right)^2} = 2\left( {y - 2} \right) is
View answer and explanation
Correct option: D
2y−3=02y - 3 = 0
Option D (2y−3=02y - 3 = 0) is correct because it represents the equation of the directrix of the given parabola, calculated using the standard form of a parabola.
305 Easy
305 of 586
Application +4 | -1
Let a focal chord of parabola y2=16x{y^2} = 16x cuts it at points (f,g)\left( {f,g} \right) and (h,k)\left( {h,k} \right) Then, f.hf.h is equal to
View answer and explanation
Correct option: B
16
Option B (16) is correct because for a focal chord of a parabola, the product of the x-coordinates of the endpoints is equal to a2a^2, where 4a4a is the coefficient of xx in the parabola's equation.
306 Easy
306 of 586
Application +4 | -1
Three normals are drawn to the parabola y2=xy^2=x through point (a,0)(a, 0). Then
View answer and explanation
Correct option: C
a>1/2a>1 / 2
Option C (a>1/2a>1 / 2) is correct because for three normals to be drawn to the parabola, the condition a>1/2a>1/2 must be satisfied.
307 Medium
307 of 586
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.
308 Easy
308 of 586
Application +4 | -1
The equation of the parabola whose directrix is y=2x−9y=2 x-9 and focus (−8,−2)(-8,-2) is
View answer and explanation
Correct option: B
x2+4y2+4xy+116x+2y+259=0x^2+4 y^2+4 x y+116 x+2 y+259=0
Option A (x2+4y2+4xy+116x+2y+259=0x^2+4 y^2+4 x y+116 x+2 y+259=0) is correct because it satisfies the definition of a parabola: the distance to the focus equals the distance to the directrix.
309 Medium
309 of 586
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.
310 Medium
310 of 586
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.
311 Medium
311 of 586
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.
312 Easy
312 of 586
Application +4 | -1
The locus of the point of intersection of perpendicular tangents to the parabola x2=4ay{x^2} = 4ay is
View answer and explanation
Correct option: B
y=−ay = - a
Option B (y=−ay = - a) is correct because the locus of the point of intersection of perpendicular tangents to a parabola is its directrix.
313 Easy
313 of 586
Application +4 | -1
The points on the parabola y2=12xy^2=12 x, whose focal distance is 4 , are
View answer and explanation
Correct option: B
(1,23),(1,−23)(1,2 \sqrt{3}),(1,-2 \sqrt{3})
Option B ((1,23),(1,−23)(1,2 \sqrt{3}),(1,-2 \sqrt{3})) is correct because substituting the focal distance into the parabola equation yields these points.
314 Easy
314 of 586
Application +4 | -1
Tangents PQP Q and PRP R are drawn to the parabola y2=4xy^2=4 x from the point P(4,5)P(4,5) then the area of the triangle PQRP Q R is
View answer and explanation
Correct option: C
272\frac{27}{2}
Option C (272\frac{27}{2}) is correct because it is the area of the triangle calculated using the formula involving the point P and the parameter 'a'.
315 Medium
315 of 586
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.

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316 Easy
316 of 586
Analyzing +4 | -1
Angle between the tangents drawn to y2=4x{y^2} = 4x at the points where it is intersected by the line y=x−1y = x - 1 is equal to
View answer and explanation
Correct option: D
π2\frac{\pi }{2}
Option D (π2\frac{\pi }{2}) is correct because the line intersects the parabola at the focus, making it a focal chord, and tangents at the ends of a focal chord are perpendicular.
317 Medium
317 of 586
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.
318 Easy
318 of 586
Application +4 | -1
The equation of the tangent at the vertex of the parabola x2+4x+2y=0{x^2} + 4x + 2y = 0 is
View answer and explanation
Correct option: C
y=2y = 2
Option C (y=2y = 2) is correct because it represents the tangent at the vertex of the given parabola after rewriting the equation in standard form.
319 Medium
319 of 586
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.
320 Easy
320 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
View answer and explanation
Correct option: D
l2=4mnl^2=4 m n
Option D (l2=4mnl^2=4 m n) is correct because it represents the condition for the line to be tangent to the parabola, derived from setting the discriminant of the resulting quadratic equation to zero.
321 Easy
321 of 586
Application +4 | -1
If the line y=mx+cy=m x+c touches the parabola y2=4a(x+a)y^2=4 a(x+a), then
View answer and explanation
Correct option: B
c=am+amc=a m+\frac{a}{m}
Option B (c=am+amc=a m+\frac{a}{m}) is correct because it represents the condition for tangency of the line to the given parabola.
322 Easy
322 of 586
Application +4 | -1
If bb and cc are the lengths of the segments of any focal chord of a parabola y2=4ax{y^2} = 4ax, then the length of the semilatusrectum is
View answer and explanation
Correct option: D
2bcb+c\frac{{2bc}}{{b + c}}
Option D (2bcb+c\frac{{2bc}}{{b + c}}) is correct because the semi-latus rectum of a parabola is the harmonic mean of the segments of any focal chord.
323 Medium
323 of 586
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.
324 Easy
324 of 586
Application +4 | -1
PQP Q is any focal chord of the parabola y2=32xy^2=32 x. The length of PQP Q can never be less than
View answer and explanation
Correct option: A
32
Option A (32) is correct because the length of the focal chord is a(t+1/t)2a(t + 1/t)^2 and the minimum value of (t+1/t)2(t + 1/t)^2 is 4, hence the minimum length is 32.
325 Easy
325 of 586
Analyzing +4 | -1
Let PQP Q be a focal chord of the parabola y2=4axy^2=4 ax such that tangents at PP and QQ meet at a point on the line y=2x+a,a>0y=2 x+a, a>0. If PQP Q subtends an angle θ\theta at the vertex of y2=4axy^2=4 a x, then tan⁡θ=\tan \theta=
View answer and explanation
Correct option: D
−253-\frac{2 \sqrt{5}}{3}
Option D (−253\frac{-2 \sqrt{5}}{3}) is correct because the tangent of the angle subtended by the focal chord PQ at the vertex is calculated considering the obtuse angle.
326 Easy
326 of 586
Application +4 | -1
The equation of the normal at the point (a4,a)\left(\frac{a}{4}, a\right) to the parabola y2=4axy^2=4 a x is
View answer and explanation
Correct option: B
4x+8y−9a=04 x+8 y-9 a=0
Option B (4x+8y−9a=04 x+8 y-9 a=0) is correct because it represents the normal equation at the given point on the parabola.
327 Easy
327 of 586
Application +4 | -1
The equation of the parabola whose axis is vertical and passes through the points (0,0),(3,0)(0,0),(3,0) and (−1,4)(-1,4) is
View answer and explanation
Correct option: A
x2−3x−y=0x^2-3 x-y=0
Option A (x2−3x−y=0x^2-3 x-y=0) is correct because it is the only equation that satisfies all three given points (0,0), (3,0), and (-1,4).
328 Medium
328 of 586
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}}.
329 Easy
329 of 586
Application +4 | -1
The point of intersection of the curves whose parametric equations are x=t2+1,y=2tx=t^2+1, y=2 t and x=2 s,y=2sx=2 \mathrm{~s}, y=\frac{2}{s}, is given by
View answer and explanation
Correct option: B
(2,2)(2,2)
Option B ((2,2)) is correct because it is the point that satisfies both parametric equations when converted to Cartesian form.
330 Hard
330 of 586
Analyzing +4 | -1
For each parabola y=x2+px+qy=x^2+p x+q, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through
View answer and explanation
Correct option: B
(0,1)(0,1)
Option B (0,1) is correct because the family of circles passing through the intersection points of the parabola and coordinate axes must pass through the point (0,1).
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