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JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

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Showing 451-480 of 586 questions

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451 Easy
451 of 586
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.
452 Easy
452 of 586
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.
453 Easy
453 of 586
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.
454 Easy
454 of 586
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.
455 Easy
455 of 586
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.
456 Easy
456 of 586
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.
457 Easy
457 of 586
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.
458 Medium
458 of 586
Analyzing +4 | -1
The parametric equation of a parabola is x=t2+1,y=2t+1.x = {t^2} + 1,y = 2t + 1. The cartesian equation of its directrix is
View answer and explanation
Correct option: A
x=0x = 0
Option A (x=0x = 0) is correct because it is the equation of the directrix of the given parabola after converting the parametric form to Cartesian form.
459 Medium
459 of 586
Application +4 | -1
The equation of the common tangent of the parabolas x2=108yx^2=108 y and y2=32xy^2=32 x, is
View answer and explanation
Correct option: B
2x+3y+36=02 x+3 y+36=0
Option B (2x+3y+36=02x + 3y + 36 = 0) is correct because it represents the common tangent to both parabolas, derived by equating the tangent equations and solving for the parameters.
460 Easy
460 of 586
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.
461 Easy
461 of 586
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.
462 Medium
462 of 586
Analyzing +4 | -1
Let f(x)=3x2−7x+cf(x)=3 x^2-7 x+c, where 'c{c}' is a variable co-efficient and x>76x > \frac{7}{6}. The value of 'c{c}' such that f(x)f(x) touches f−1(x)f^{-1}(x) is.....
View answer and explanation
Correct option: C
163\frac{16}{3}
Option C (163\frac{{16}}{{3}}) is correct because f(x) touches f⁻¹(x) when f(x) = x, and solving the resulting quadratic equation with equal roots yields c = 16/3.
463 Easy
463 of 586
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.
464 Easy
464 of 586
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.
465 Medium
465 of 586
Application +4 | -1
If line y=2x+cy=2 x+c is a normal to the parabola y2=4xy^2=4 x at the point (t2,2t)\left(t^2, 2 t\right) then
View answer and explanation
Correct option: A
c=−12,t=−2c=-12, t=-2
Option A (c=-12, t=-2) is correct because substituting the normal equation into the given line equation and comparing coefficients yields these values for c and t.

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466 Easy
466 of 586
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).
467 Medium
467 of 586
Application +4 | -1
The length of the focal chord of parabola y2=4axy^2=4 a x that makes an angle α\alpha with the xx-axis, is
View answer and explanation
Correct option: C
4acosec⁡2α4 a \operatorname{cosec}^2 \alpha
Option C (4acosec⁡2α4 a \operatorname{cosec}^2 \alpha) is correct because the length of the focal chord is derived using the angle it makes with the x-axis and the property t1t2=−1t_1t_2 = -1.
468 Easy
468 of 586
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.
469 Easy
469 of 586
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.
470 Medium
470 of 586
Application +4 | -1
If mm be the slope of common tangents to y=x2−x+1y=x^2-x+1 and y=x2−3x+10y=x^2-3 x+10, then m=m=
View answer and explanation
Correct option: B
7
Option B (7) is correct because by setting the condition for a common tangent to both parabolas, we can solve for the slope mm, which equals 7.
471 Medium
471 of 586
Application +4 | -1
If y1,  y2{y_1},\;{y_2} and y3{y_3} are the ordinates of the vertices of a triangle inscribed in the parabola y2=4ax{y^2} = 4ax, then its area is
View answer and explanation
Correct option: C
18a∣(y1−y2)(y2−y3)(y3−y1)∣\frac{1}{{8a}}\left| {\left( {{y_1} - {y_2}} \right)\left( {{y_2} - {y_3}} \right)\left( {{y_3} - {y_1}} \right)} \right|
Option C (\frac{1}{{8a}}\left| {\left( {{{y_1} - {y_2}} \right)\left( {{{y_2} - {y_3}} \right)\left( {{{y_3} - {y_1}} \right)} \right|\frac{1}{{8a}}\left| {\left( {{{y_1} - {y_2}} \right)\left( {{{y_2} - {y_3}} \right)\left( {{{y_3} - {y_1}} \right)} \right|) is correct because it represents the area of a triangle inscribed in a parabola in terms of the ordinates of the vertices.
472 Easy
472 of 586
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.
473 Easy
473 of 586
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.
474 Easy
474 of 586
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.
475 Easy
475 of 586
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.
476 Easy
476 of 586
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.
477 Hard
477 of 586
Analyzing +4 | -1
If the normals from any point to the parabola x2=4yx^2=4 y cuts the line y=2y=2 in points whose abscissae are in A.P., then the slopes of the tangents at the 3 conormal points are in
View answer and explanation
Correct option: B
GPG P
Option B (G P) is correct because the slopes of the tangents at the three conormal points are in Geometric Progression.
478 Medium
478 of 586
Application +4 | -1
The area of triangle formed inside the parabola y2=4xy^2=4 x and whose ordinates of vertices are 1,2 and 4 is
View answer and explanation
Correct option: D
3/43 / 4
Option D (3/43/4) is correct because the area of the triangle is calculated using the determinant formula with the given ordinates.
479 Easy
479 of 586
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.
480 Easy
480 of 586
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.
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