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JEE • MATHEMATICS • PARABOLA

Parabola practice · Mixed difficulty

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Showing 331-360 of 586 questions

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331 Easy
331 of 586
Application +4 | -1
The tangent to the parabola y2=4x{y^2} = 4x at the point where it intersects the circle x2+y2=5{x^2} + {y^2} = 5 in the first quadrant, passes through the point
View answer and explanation
Correct option: B
(34,74)\left( {\frac{3}{4},\frac{7}{4}} \right)
Option B ((34,74)\left( {\frac{3}{4},\frac{7}{4}} \right)) is correct because the tangent to the parabola at the point of intersection with the circle in the first quadrant passes through this point.
332 Medium
332 of 586
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.
333 Easy
333 of 586
Application +4 | -1
The vertex of a parabola is the point (a,b)(a, b) and Latusrectum is of length l.l. If the axis of the parabola is along the positive direction of yy-axis, then its equation is
View answer and explanation
Correct option: B
(x−a)2=l2(2y−2b)(x-a)^2=\frac{l}{2}(2 y-2 b)
Option B ((x−a)2=l2(2y−2b)(x-a)^2=\frac{l}{2}(2 y-2 b)) is correct because it represents a parabola with vertex (a,b), latus rectum l, and axis along the positive y-axis.
334 Easy
334 of 586
Application +4 | -1
If the focus of a parabola is at (0,−3)(0,-3) and its directrix is y=3y=3, then its equation is
View answer and explanation
Correct option: A
x2=−12yx^2=-12 y
Option A (x2=−12yx^2=-12 y) is correct because it satisfies the definition of a parabola with focus (0,-3) and directrix y=3.
335 Easy
335 of 586
Application +4 | -1
If P(t1),Q(t2)P\left(t_1\right), Q\left(t_2\right) and R(t3)R\left(t_3\right) are three points on the parabola y2=4axy^2=4 a x, then area of ΔPQR=\Delta P Q R=
View answer and explanation
Correct option: B
a2∣(t1−t2)(t2−t3)(t3−t1)∣a^2\left|\left(t_1-t_2\right)\left(t_2-t_3\right)\left(t_3-t_1\right)\right|
Option A (a∣(t1−t2)(t2−t3)(t3−t1)∣a|(t_1-t_2)(t_2-t_3)(t_3-t_1)|) is correct because it represents the area of the triangle formed by three points on the parabola in terms of their parameters.
336 Easy
336 of 586
Application +4 | -1
The equation 16x2+y2+8xy−74x−78y+212=016 x^2+y^2+8 x y-74 x-78 y+212=0 represents
View answer and explanation
Correct option: B
A parabola
Option B (A parabola) is correct because the discriminant Δ≠0\Delta \neq 0 and h2=abh^2 = ab, which are the conditions for a parabola.
337 Medium
337 of 586
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).
338 Easy
338 of 586
Application +4 | -1
The length of the latusrectum of the parabola whose focus is (3,3)\left( {3,3} \right) and directrix is 3x−4y−2=0,  3x - 4y - 2 = 0,\; is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the length of the latus rectum is twice the distance between the focus and the directrix.
339 Medium
339 of 586
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.
340 Medium
340 of 586
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.
341 Medium
341 of 586
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).
342 Easy
342 of 586
Analyzing +4 | -1
Let AA be a point on the xx-axis. Common tangents are drawn from AA to the curves x2+y2=8x^2+y^2=8 and y2=16xy^2=16 x. If one of these tangents touches the two curves at QQ and RR, then (QR)2(Q R)^2 is equal to
View answer and explanation
Correct option: B
72
Option B is correct because the square of the distance between the points of tangency QQ and RR is 72.
343 Medium
343 of 586
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.
344 Easy
344 of 586
Application +4 | -1
If the tangent to the parabola y2=ax{y^2} = ax makes an angle of 450{45^0} with x−axisx - axis, then the point of contact is
View answer and explanation
Correct option: D
(a4,a2)\left( {\frac{a}{4},\frac{a}{2}} \right)
Option D (a4,a2)\left( {\frac{a}{4},\frac{a}{2}} \right) is correct because it is the point of contact where the tangent to the parabola makes a 45-degree angle with the x-axis.
345 Easy
345 of 586
Application +4 | -1
The straight line x+y=kx + y = k touches the parabola y=x−x2,y = x - {x^2}, if k =
View answer and explanation
Correct option: C
1
Option C (1) is correct because the line touches the parabola when the discriminant of the resulting quadratic equation is zero, leading to k=1.

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346 Hard
346 of 586
Understanding +4 | -1
The locus of the foot of the perpendicular from the focus upon a tangent to the parabola y2=4  ax{y^2} = 4\;ax is
View answer and explanation
Correct option: B
Tangent at the vertex
Option B (Tangent at the vertex) is correct because the locus of the foot of the perpendicular from the focus upon a tangent to the parabola is the tangent at the vertex.
347 Easy
347 of 586
Analyzing +4 | -1
The number of distinct normal that can be drawn from (11/4,1/4)\left( {11/4,1/4} \right) to the parabola y2=4x,{y^2} = 4x, is
View answer and explanation
Correct option: A
3
Option A (3) is correct because there are three distinct normal lines that can be drawn from the given point to the parabola.
348 Easy
348 of 586
Application +4 | -1
A tangent is drawn to the parabola y2=6x{y^2} = 6x, which is perpendicular to the line 2x+y=1.2x + y = 1. Which of the following points does not lie on it?
View answer and explanation
Correct option: C
(5,4)\left( {5,4} \right)
Option C (5,4) is correct because it does not satisfy the equation of the tangent to the parabola.
349 Easy
349 of 586
Application +4 | -1
The number of distinct normals that can be drawn to parabola y2=16x{y^2} = 16x from the point (2,0)\left( {2,0} \right), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one real value of the slope mm satisfies the condition for the normal to pass through the given point.
350 Easy
350 of 586
Application +4 | -1
If the normal to the parabola y2=4axy^2=4 a x at the point (at 2,2at)\left.{ }^2, 2 a t\right) cuts the parabola again at (aT2,2aT)\left({aT}^2, 2 {aT}\right), then
View answer and explanation
Correct option: D
T2≥8T^2 \geq 8
Option D (T2≥8T^2 \geq 8) is correct because the condition for the normal to cut the parabola again is derived from the quadratic equation having real roots, leading to the inequality T2≥8T^2 \geq 8.
351 Easy
351 of 586
Application +4 | -1
The focus of the parabola 4y2−6x−4y=54 y^2-6 x-4 y=5 is
View answer and explanation
Correct option: B
(−5/8,1/2)(-5 / 8,1 / 2)
Option B ((-5 / 8, 1 / 2)) is correct because the focus of the parabola is (-5/8, 1/2) after rewriting the equation in standard form.
352 Easy
352 of 586
Analyzing +4 | -1
If the normals of the parabola y2=4xy^2=4 x drawn at the end points of its latusrectum are tangents to the circle (x−3)2+(y+2)2=r2(x-3)^2+(y+2)^2=r^2, then the value of r2r^2, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the value of r2r^2 is found to be 2 by substituting the center of the circle into the normal equations and solving.
353 Medium
353 of 586
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).
354 Easy
354 of 586
Application +4 | -1
If two tangents drawn from a point PP to the parabola y2=16(x−3){y^2} = 16(x - 3) are at right angles, then the locus of point PP is
View answer and explanation
Correct option: B
x+1=0x + 1 = 0
Option B (x+1=0x + 1 = 0) is correct because the locus of the point P from which two tangents drawn to the parabola are at right angles is the directrix of the parabola.
355 Easy
355 of 586
Analyzing +4 | -1
The triangle formed by the tangents to the parabola y2=4axy^2=4 a x at the ends of the latusrectum and the double ordinate through the focus, is
View answer and explanation
Correct option: C
right-angled isosceles
Option C (right-angled isosceles) is correct because the tangents at the ends of a focal chord intersect at right angles, and the triangle is isosceles due to symmetry.
356 Easy
356 of 586
Application +4 | -1
Equation of the common tangent to y2=4xy^2=4 x and x2=32yx^2=32 y is
View answer and explanation
Correct option: A
x+2y+4=0x+2 y+4=0
Option A (x+2y+4=0x+2y+4=0) is correct because it represents the common tangent found by solving the equations of tangents to both curves and equating them.
357 Easy
357 of 586
Analyzing +4 | -1
The inclination of the normal with positive direction of xx-axis, drawn at another end of a normal to the parabola y2=4xy^2=4 x is always.
View answer and explanation
Correct option: C
more than 60∘60^{\circ}
Option C (more than 60∘60^{\circ}) is correct because the inclination of the normal is always greater than 60∘60^{\circ} as derived using AM >= GM.
358 Medium
358 of 586
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.
359 Easy
359 of 586
Analyzing +4 | -1
If a≠0a \neq 0 and the line 2bx+3cy+4d=02 b x+3 c y+4 d=0 passes through the points of intersection of the parabolas y2=4axy^2=4 a x and x2=4ayx^2=4 a y, then
View answer and explanation
Correct option: D
d2+(2b+3c)2=0d^2+(2 b+3 c)^2=0
Option B ((a/2,0)(a / 2,0)) is incorrect. The correct answer is C ((a,0)(a, 0)) because the locus of the midpoints is another parabola whose focus is at (a,0).
360 Easy
360 of 586
Application +4 | -1
Equation of a normal to the parabola y2=32xy^2=32 x passing through its focus is
View answer and explanation
Correct option: B
y=0y=0
Option B (y=0y=0) is correct because it represents a normal to the parabola that passes through its focus, derived from the normal equation and the focus coordinates.
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