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JEE • MATHEMATICS • PARABOLA

Parabola practice · Easy

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241 Easy
241 of 356
Application +4 | -1
If (0,6)\left( {0,6} \right) and (0,3)\left( {0,3} \right) are respectively the vertex and focus of a parabola, then its equation is
View answer and explanation
Correct option: A
x2+12y=72{x^2} + 12y = 72
Option A (x2+12y=72{x^2} + 12y = 72) is correct because it represents a parabola with the given vertex and focus.
242 Easy
242 of 356
Application +4 | -1
The normal to the parabola y2=8x{y^2} = 8x at the point (2,  4)\left( {2,\;4} \right) meets the parabola again at the point
View answer and explanation
Correct option: D
(18,  −12)\left( {18,\; - 12} \right)
Option D ((18,  −12)\left( {18,\; - 12} \right)) is correct because it's the point where the normal at (2, 4) intersects the parabola again, calculated using the parametric form of the parabola.
243 Easy
243 of 356
Application +4 | -1
The locus of the poles of tangents to the parabola y2=4axy^2=4 a x with respect to the parabola y2=4bxy^2=4 b x is
View answer and explanation
Correct option: B
a parabola
Option B (a parabola) is correct because the locus of the poles is found to be a parabola by relating the tangent equation to the polar equation.
244 Easy
244 of 356
Application +4 | -1
The equation of the parabola with focus (0,0)\left( {0,0} \right) and directrix x+y=4x + y = 4, is
View answer and explanation
Correct option: A
x2+y2−2xy+8x+8y−16=0{x^2} + {y^2} - 2xy + 8x + 8y - 16 = 0
Option A (x2+y2−2xy+8x+8y−16=0{x^2} + {y^2} - 2xy + 8x + 8y - 16 = 0) is correct because it is found by using the definition of a parabola: the distance to the focus equals the distance to the directrix.
245 Easy
245 of 356
Application +4 | -1
The point of contact of the tangent 18x−6y+1=018 x-6 y+1=0 to the parabola y2=2xy^2=2 x is
View answer and explanation
Correct option: D
(118,13)\left(\frac{1}{18}, \frac{1}{3}\right)
Option D (118,13)\left(\frac{{1}}{{18}}, \frac{{1}}{{3}}\right) is correct because it represents the point where the given tangent touches the parabola, found by comparing coefficients.
246 Easy
246 of 356
Application +4 | -1
The equation of the parabola with vertex at the origin and directrix y=2y = 2 is
View answer and explanation
Correct option: D
x2=−8y{x^2} = - 8y
Option D (x2=−8y{x^2} = - 8y) is correct because with the vertex at the origin and the directrix at y=2y = 2, the parabola opens downwards, and the equation is of the form x2=−4ay{x^2} = -4ay, where a=2a = 2.
247 Easy
247 of 356
Application +4 | -1
The tangents to a parabola at its vertex AA and any point PP meet at QQ. If SS be the focus of the parabola, then SP,SQS P, S Q, SAS A are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because SQ2=SP×SASQ^2 = SP \times SA, which is the condition for SP,SQ,SASP, SQ, SA to be in Geometric Progression.
248 Easy
248 of 356
Application +4 | -1
The points on the parabola y2=36xy^2=36 x, whose ordinate is three times the abscissa, are
View answer and explanation
Correct option: A
(0,0),(4,12)(0,0),(4,12)
Option A ((0,0),(4,12)) is correct because these points satisfy both the parabola's equation and the condition that the ordinate is three times the abscissa.
249 Easy
249 of 356
Application +4 | -1
If the normal chord of the parabola y2=4xy^2=4 x makes an angle of 45∘45^{\circ} with the axis of the parabola, then its length, is
View answer and explanation
Correct option: B
828 \sqrt{2}
Option B (828 \sqrt{2}) is correct because by finding the coordinates of the points where the normal chord intersects the parabola and calculating the distance between them.
250 Easy
250 of 356
Application +4 | -1
The equation of the common tangent to the circle (x−3)2+y2=9(x-3)^2+y^2=9 and the parabola y2=4xy^2=4 x above the xx-axis, 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 by finding the tangent to the parabola and then applying the condition for tangency to the circle, we arrive at this equation.
251 Easy
251 of 356
Application +4 | -1
The two parabolas x2=4y{x^2} = 4y and y2=4x{y^2} = 4x meet in two distinct points. One of these is the origin and the other is
View answer and explanation
Correct option: C
(4,4)\left( {4,4} \right)
Option C ((4,4)) is correct because it is the non-origin intersection point found by solving the system of equations for the two parabolas.
252 Easy
252 of 356
Application +4 | -1
The equation of parabola whose vertex and focus are (0,4)(0,4) and (0,2)(0,2), respectively, is
View answer and explanation
Correct option: C
x2+8y=32x^2+8 y=32
Option C (x2+8y=32x^2+8 y=32) is correct because it represents a parabola opening downwards with the given vertex and focus.
253 Easy
253 of 356
Analyzing +4 | -1
If a line along a chord of the circle 4x2+4y2+120x+675=04{x^2} + 4{y^2} + 120x + 675 = 0 passes through the point (-30, 0) and is tangent to the parabola y2=30x{y^2} = 30x, then the length of this chord is
View answer and explanation
Correct option: D
353\sqrt 5
The correct answer is D (353\sqrt 5) because it is the length of the chord calculated using the distance from the center of the circle to the tangent line.
254 Easy
254 of 356
Application +4 | -1
The focus of the parabola x2+2y+6x=0{x^2} + 2y + 6x = 0 is
View answer and explanation
Correct option: A
(−3,4)\left( { - 3,4} \right)
Option A ((-3,4)) is correct because it is found by completing the square and transforming the equation to standard form to identify the focus.
255 Easy
255 of 356
Application +4 | -1
The vertex of the parabola whose focus is (−1,1)(-1,1) and directrix is 4x+3y−24=04 x+3 y-24=0 is
View answer and explanation
Correct option: D
(1,5/2)(1 , 5 / 2)
Option D ((1,5/2)(1 , 5 / 2)) is correct because the vertex is the midpoint between the focus and the foot of the perpendicular from the focus to the directrix.

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256 Easy
256 of 356
Application +4 | -1
The point of the contact of the tangent to the parabola y2=4axy^2=4 a x, which makes an angle of 60∘60^{\circ} with xx-axis, is
View answer and explanation
Correct option: A
(a3,2a3)\left(\frac{a}{3}, \frac{2 a}{\sqrt{3}}\right)
Option A ((a3,2a3)\left(\frac{a}{3}, \frac{2 a}{\sqrt{3}}\right)) is correct because it represents the point of contact of the tangent to the parabola y2=4axy^2=4 a x making an angle of 60∘60^{\circ} with the x-axis.
257 Easy
257 of 356
Application +4 | -1
Let (x,y)(x, y) be any point on the parabola y2=4xy^2=4 x. Let PP be the point that divides the line segment from (0,0)(0,0) to (x,y)(x, y) in the ratio 1:3. Then the locus of PP is
View answer and explanation
Correct option: C
y2=xy^2=x
Option C (y2=xy^2=x) is correct because it is the locus of the point P that divides the line segment from (0,0) to (x,y) in the ratio 1:3.
258 Easy
258 of 356
Application +4 | -1
The ends of latus rectum of parabola x2+8y=0x^2+8 y=0 are
View answer and explanation
Correct option: C
(−4,−2)(-4,-2) and (4,−2)(4,-2)
Option C ((−4,−2)(-4,-2) and (4,−2)(4,-2)) is correct because these points satisfy the parabola equation and are at a distance of 2a from the focus.
259 Easy
259 of 356
Application +4 | -1
If VV and SS are respectively the vertex and focus of the parabola y2+6y+2x+5=0y^2+6 y+2 x+5=0, then SV=S V=
View answer and explanation
Correct option: B
1/21 / 2
Option B (1/21/2) is correct because the distance between the vertex and focus is one-fourth of the latus rectum length.
260 Easy
260 of 356
Application +4 | -1
The length of the latusrectum of the parabola 169{(x−1)2+(y−3)2}=(5x−12y+7)2169\left\{(x-1)^2+(y-3)^2\right\}=(5 x-12 y+7)^2, is
View answer and explanation
Correct option: D
4813\frac{48}{13}
Option D (4813\frac{{48}}{{13}}) is correct because the length of the latus rectum is twice the perpendicular distance from the focus to the directrix.
261 Easy
261 of 356
Application +4 | -1
If the three normals drawn to the parabola, y2=2x{y^2} = 2x pass through the point (a, 0)\left( {a,\,0} \right) a≠0a \ne 0, then ‘aa’ must be greater than:
View answer and explanation
Correct option: A
1
Option A (1) is correct because the condition for three normals to pass through a point on the axis of the parabola requires 'a' to be greater than 1.
262 Easy
262 of 356
Application +4 | -1
The vertex of the parabola x2+8x+12y+4=0x^2+8 x+12 y+4=0, is
View answer and explanation
Correct option: A
(−4,1)(-4,1)
Option A ((-4,1)) is correct because completing the square and rewriting the equation reveals the vertex form of the parabola.
263 Easy
263 of 356
Application +4 | -1
The normal at the point (bt12,2bt1)\left(b t_1^2, 2 b t_1\right) on a parabola meets the parabola again in the point (bt22,2bt2)\left(b t_2^2, 2 b t_2\right), then
View answer and explanation
Correct option: B
t2=−t1−2t1t_2=-t_1-\frac{2}{t_1}
Option B (t2=−t1−2t1t_2=-t_1-\frac{2}{t_1}) is correct because it represents the relationship between the parameters of the points where the normal intersects the parabola.
264 Easy
264 of 356
Application +4 | -1
The equation of parabola whose vertex and focus lie on the axis of xx at distances aa and a1a_1 from the origin respectively, is
View answer and explanation
Correct option: B
y2=4(a1−a)(x−a)y^2=4\left(a_1-a\right)(x-a)
Option B (y2=4(a1−a)(x−a)y^2=4(a_1-a)(x-a)) is correct because it represents the equation of a parabola with vertex at (a,0) and focus at (a1,0).
265 Easy
265 of 356
Application +4 | -1
If PSQ is a focal chord of the parabola y2=8xy^2=8 x such that SP=6S P=6, then SQ=S Q=
View answer and explanation
Correct option: C
3
Option C (3) is correct because the semi-latus rectum is the harmonic mean of SP and SQ, and solving for SQ gives the correct value.
266 Easy
266 of 356
Application +4 | -1
Tangent to the parabola y=x2+6y=x^2+6 at (1,7)(1,7) touches the circle x2+y2+16x+12y+c=0x^2+y^2+16 x+12 y+c=0 at the point
View answer and explanation
Correct option: C
(−6,−7)(-6,-7)
Option C ((-6,-7)) is correct because it is the point of tangency between the line and the circle.
267 Easy
267 of 356
Application +4 | -1
If y=2x+ky = 2x + k is a tangent to the curve x2=4y,{x^2} = 4y, then kk is equal to
View answer and explanation
Correct option: C
−4- 4
Option C (-4) is correct because substituting the tangent equation into the parabola equation and solving for the tangency condition yields k = -4.
268 Easy
268 of 356
Application +4 | -1
If the line x+y−1=0x+y-1=0 touches the parabola y2=kxy^2=k x, then the value of kk, is
View answer and explanation
Correct option: B
-4
Option B (-4) is correct because substituting the line equation into the parabola equation and applying the tangency condition (discriminant equals zero) yields k = -4.
269 Easy
269 of 356
Application +4 | -1
If a focal chord of the parabola y2=ax{y^2} = ax is 2x−y−8=0,2x - y - 8 = 0, then the equation of the directrix is
View answer and explanation
Correct option: A
x+4=0x + 4 = 0
Option A (x+4=0x + 4 = 0) is correct because the focal chord passes through the focus, allowing us to determine 'a' and subsequently the directrix.
270 Easy
270 of 356
Application +4 | -1
The length of latus rectum of the parabola 4y2+2x−20y4 y^2+2 x-20 y +17=0+17=0 is
View answer and explanation
Correct option: C
1/21 / 2
Option C (1/21 / 2) is correct because the latus rectum of the parabola is 1/2 after rewriting the equation in standard form.
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