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JEE • MATHEMATICS • PARABOLA

Parabola practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 271-300 of 356 questions

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271 Easy
271 of 356
Analyzing +4 | -1
The chord ABAB of the parabola y2=4ax{y^2} = 4ax cuts the axis of the parabola at CC. If A=(at12,  2at1),  B=(at22,  2at2)A = \left( {at_1^2,\;2a{t_1}} \right),\;B = \left( {at_2^2,\;2a{t_2}} \right) and AC:AB=1:3,AC:AB = 1:3, then
View answer and explanation
Correct option: B
t2+2t1=0{t_2} + 2{t_1} = 0
Option B (t2+2t1=0{t_2} + 2{t_1} = 0) is correct because it satisfies the condition AC:AB = 1:3, where C lies on the x-axis.
272 Easy
272 of 356
Application +4 | -1
The equation of parabola whose vertex (0,0)(0,0) and focus (3,0)(3,0) is y2=4axy^2=4 a x. 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 parabola's equation is 3, derived from the focus coordinates.
273 Easy
273 of 356
Application +4 | -1
If the vertex of a parabola is the point (−3,0)\left( { - 3,0} \right) and the directrix is the line x+5=0,x + 5 = 0, then its equation is
View answer and explanation
Correct option: A
y2=8(x+3){y^2} = 8\left( {x + 3} \right)
Option A is correct because the equation of the parabola is derived from the given vertex and directrix using the definition of a parabola.
274 Easy
274 of 356
Application +4 | -1
In the parabola y2=4axy^2=4 a x, the length of the chord through the vertex and inclined to the axis at π/4\pi / 4, is
View answer and explanation
Correct option: A
42a4 \sqrt{2} a
Option A (42a4 \sqrt{{2}} a) is correct because it represents the length of the chord calculated using the parametric form and the given angle.
275 Easy
275 of 356
Application +4 | -1
The directrix of the parabola x2−4x−8y+12=0{x^2} - 4x - 8y + 12 = 0 is
View answer and explanation
Correct option: C
y=−1y = - 1
Option C (y=−1y = - 1) is correct because it represents the directrix of the given parabola, found by rewriting the equation in standard form and identifying the value of A.
276 Easy
276 of 356
Application +4 | -1
The locus of the point of intersection of tangents to the parabola y2=4axy^2=4 a x which meet at an angle α\alpha, is
View answer and explanation
Correct option: A
(x+a)2tan⁡2α=y2−4ax(x+a)^2 \tan ^2 \alpha=y^2-4 a x
Option A ((x+a)2tan⁡2α=y2−4ax(x+a)^2 \tan ^2 \alpha=y^2-4 a x) is correct because it represents the locus derived from the condition that the tangents intersect at an angle α\alpha.
277 Easy
277 of 356
Application +4 | -1
The length of the latusrectum of the parabola x=ay2+by+cx=a y^2+b y+c, is
View answer and explanation
Correct option: C
1/a1 / a
Option C (1 / a) is correct because the length of the latus rectum is the reciprocal of the coefficient of the y² term when x is expressed in terms of y.
278 Easy
278 of 356
Application +4 | -1
The focus of the parabola y2=4y−4xy^2=4 y-4 x is
View answer and explanation
Correct option: A
(0,2)(0,2)
Option A ((0,2)) is correct because by completing the square, the equation can be rewritten in standard form, revealing the vertex and thus the focus.
279 Easy
279 of 356
Application +4 | -1
If the chord y=mx+cy = mx + c subtends a right angle at the vertex of the parabola y2=4  ax,{y^2} = 4\;ax, then the value of cc is
View answer and explanation
Correct option: A
−4 am- 4\,\mathrm{am}
Option A (−4 am- 4\,\mathrm{am}) is correct because the condition for the chord to subtend a right angle at the vertex of the parabola is used.
280 Easy
280 of 356
Application +4 | -1
The equation of the chord of contact of tangents from (2,5)(2,5) to the parabola y2=8xy^2=8 x, is
View answer and explanation
Correct option: B
4x−5y+8=04 x-5 y+8=0
Option D (4x+5y−8=04 x+5 y-8=0) is correct because it is the equation of the chord of contact found by using the formula T = 0, where T is obtained by replacing y2y^2 with yy1yy_1 and xx with (x+x1)/2(x+x_1)/2.
281 Easy
281 of 356
Application +4 | -1
If the point (a,−1)(a,-1) is exterior to both the parabolas y2=∣x∣y^2=|x|, then
View answer and explanation
Correct option: B
a∈(−1,1)a \in(-1,1)
Option B (a∈(−1,1)a \in(-1,1)) is correct because the point must satisfy the exterior condition for both parabolas, leading to the intersection of the inequalities.
282 Easy
282 of 356
Application +4 | -1
The directrix of the parabola y2+4x+3=0{y^2} + 4x + 3 = 0 is
View answer and explanation
Correct option: D
x−14=0x - \frac{1}{4} = 0
Option D (x−14=0x - \frac{1}{4} = 0) is correct because it represents the directrix of the given parabola.
283 Easy
283 of 356
Application +4 | -1
If aa and cc are the segments of a focal chord of a parabola and bb the semi-latus rectum, then
View answer and explanation
Correct option: C
a,b,ca, b, c are in HP.
Option D ((1,−2)(1,-2)) is correct because the normal's slope is ±1\pm 1 when it makes equal angles with the axes, leading to the point (1, -2).
284 Easy
284 of 356
Analyzing +4 | -1
If three distinct normals are drawn from (2k,0)(2 k, 0) to the parabola y2=4xy^2=4 x such that one of them is xx-axis and other two are perpendicular, then k=k=
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C (32\frac{3}{2}) is correct because the value of k is found by ensuring that two of the three normals drawn from (2k, 0) to the parabola are perpendicular, with one being the x-axis.
285 Easy
285 of 356
Application +4 | -1
Normals at two points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) of the parabola y2=4xy^2=4 x meet again on the parabola, where x1+x2=4x_1+x_2=4, then ∣y1+y2∣\left|y_1+y_2\right| is equal to
View answer and explanation
Correct option: C
424 \sqrt{2}
Option C (424 \sqrt{2}) is correct because it is derived from the given condition x1+x2=4x_1 + x_2 = 4 and the properties of the parabola y2=4xy^2 = 4x.

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286 Easy
286 of 356
Application +4 | -1
If the line x=my+kx=m y+k touches the parabola x2=4ayx^2=4 a y, then what is the value of kk ?
View answer and explanation
Correct option: A
a/ma / m
Option A (a/ma / m) is correct because it is the condition for the line to be tangent to the parabola.
287 Easy
287 of 356
Application +4 | -1
The length of the shortest normal chord of the parabola y2=4axy^2=4 a x is
View answer and explanation
Correct option: A
2a272 a \sqrt{27}
Option A (2a272 a \sqrt{27}) is correct because it represents the length of the shortest normal chord of the parabola.
288 Easy
288 of 356
Application +4 | -1
The locus of the poles w.r.t. the parabola y2=4axy^2=4 a x of the tangents to the rectangular hyperbola x2−y2=a2x^2-y^2=a^2 is the ellipse
View answer and explanation
Correct option: C
4x2+y2=4a24 x^2+y^2=4 a^2
Option D is incorrect, the correct answer is 4x2+y2=4a24x^2 + y^2 = 4a^2, which is not listed.
289 Easy
289 of 356
Application +4 | -1
The pole of the line lx+my+n=0l x+m y+n=0 with respect to the parabola y2=4axy^2=4 a x, is
View answer and explanation
Correct option: A
(nl,−2aml)\left(\frac{n}{l},-\frac{2 a m}{l}\right)
Option A is correct because the pole of the line can be found by comparing the equation of the polar with respect to the parabola with the given line.
290 Easy
290 of 356
Application +4 | -1
If the normal chord at a point ' tt ' on the parabola y2=4axy^2=4 a x subtends a right angle at the vertex, then a value of tt is
View answer and explanation
Correct option: C
2\sqrt{2}
Option D (y=x+2y=x+2) is correct because it represents a common tangent to both the curves, derived from the tangency conditions.
291 Easy
291 of 356
Application +4 | -1
The equation of the normal to the parabola y2=4xy^2=4 x, parallel to the line y=2x−5y=2 x-5, is
View answer and explanation
Correct option: B
y=2x−12y=2 x-12
Option B (y=2x−12y=2 x-12) is correct because by using the normal equation to the parabola and comparing it with the given parallel line, we find the equation of the normal.
292 Easy
292 of 356
Application +4 | -1
The locus of the mid point of the line joining the focus and any point on the parabola y2=4ax{y^2} = 4ax is a parabola with the equation of directrix as
View answer and explanation
Correct option: C
x=0x = 0
Option C (x=0x = 0) is correct because the directrix of the new parabola is x=0x=0.
293 Easy
293 of 356
Application +4 | -1
The locus of the point of intersection of two tangents to the parabola y2=4axy^2=4ax which include at an angle of 45∘45^{\circ}, is
View answer and explanation
Correct option: A
y2−4ax=(a+x)2y^2-4 a x=(a+x)^2
Option A (y2−4ax=(a+x)2y^2-4 a x=(a+x)^2) is correct because it is derived by substituting α=45∘\alpha = 45^{\circ} into the general equation for the locus of intersection of tangents at an angle α\alpha.
294 Easy
294 of 356
Application +4 | -1
The value of θ∈(0,π/2)\theta \in(0, \pi / 2) for which the line y=xcos⁡θ+4cos⁡3θ−14cos⁡θ−1y=x \cos \theta+4 \cos ^3 \theta-14 \cos \theta-1 is a normal to the parabola y2=16xy^2=16 x, is
View answer and explanation
Correct option: C
π/9\pi / 9
Option C (π/9\pi / 9) is correct because it is the value of θ\theta that satisfies the condition for the given line to be a normal to the parabola.
295 Easy
295 of 356
Application +4 | -1
Length of the normal chord of the parabola, y2=4xy^2=4 x, which makes an angle of π4\frac{\pi}{4} with the axis of xx is:
View answer and explanation
Correct option: B
828 \sqrt{2}
Option B (828 \sqrt{2}) is correct because it is the calculated length of the normal chord using the given angle and parabola equation.
296 Easy
296 of 356
Analyzing +4 | -1
The parabolas : ax2+2bx+cy=0a x^2+2 b x+c y=0 and dx2+2ex+fy=0d x^2+2 e x+f y=0 intersect on the line y=1y=1.

If a,b,c,d,e,fa, b, c, d, e, f are positive real numbers and a,ba, b, cc are in G.P., then
View answer and explanation
Correct option: A
da,eb,fc\frac{d}{a}, \frac{e}{b}, \frac{f}{c} are in A.P.
Option A (da,eb,fc\frac{{d}}{{a}}, \frac{{e}}{{b}}, \frac{{f}}{{c}} are in A.P.) is correct because by finding the intersection point and using the G.P. condition, we can show that the ratios are in arithmetic progression.
297 Easy
297 of 356
Application +4 | -1
If P(at12,2at1)P\left(a t_1^2, 2 a t_1\right) and Q(at22,2at2)Q\left(a t_2^2, 2 a t_2\right) are two points on the parabola y2=4axy^2=4 a x, then the area of the triangle formed by the tangents at PP and QQ and the chord PQP Q, is
View answer and explanation
Correct option: A
12a2∣t1−t2∣3\frac{1}{2} a^2\left|t_1-t_2\right|^3
Option A (12a2∣t1−t2∣3\frac{1}{2} a^2\left|t_1-t_2\right|^3) is correct because it represents the area of the triangle formed by the tangents and the chord, derived from the determinant formula.
298 Easy
298 of 356
Application +4 | -1
The equation to the line touching both the parabolas y2=4x{y^2} = 4x and x2=−32y,{x^2} = - 32y, is
View answer and explanation
Correct option: D
x−2y+4=0x - 2y + 4 = 0
Option C (x−2y−4=0x - 2y - 4 = 0) is incorrect. Option D (x−2y+4=0x - 2y + 4 = 0) is correct because this line is tangent to both parabolas y2=4x{y^2} = 4x and x2=−32y{x^2} = - 32y.
299 Easy
299 of 356
Analyzing +4 | -1
If a,b,ca, b, c are distinct positive real numbers such that the parabolas y2=4axy^2=4 a x and y2=4c(x−b)y^2=4 c(x-b) will have a common normal, then
View answer and explanation
Correct option: B
ba−c>2\frac{b}{a-c}>2
Option B (ba−c>2\frac{b}{a-c}>2) is correct because it represents the condition for the existence of a common normal to the two parabolas.
300 Easy
300 of 356
Application +4 | -1
If the line 2x−1=02 x-1=0 is the directrix of the parabola y2−kx+6=0y^2-k x+6=0 then the positive values of kk is
View answer and explanation
Correct option: A
4
Option A (4) is correct because by equating the given directrix with the directrix derived from the parabola equation, we find the positive value of k to be 4.
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