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JEE • MATHEMATICS • PARABOLA

Parabola practice · Medium

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Showing 151-180 of 211 questions

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151 Medium
151 of 211
Application +4 | -1
If lx+my+n=0l x+m y+n=0 is tangent to the parabola x2=yx^2=y, then the condition of tangency is
View answer and explanation
Correct option: D
l2=4mnl^2=4 \mathrm{mn}
Option D (l2=4mnl^2=4 \mathrm{mn}) is correct because this is the condition derived when the discriminant of the quadratic equation formed by the intersection of the line and parabola is set to zero.
152 Medium
152 of 211
Application +4 | -1
The locus of the point of intersection of tangents to the parabola y2=4(x+1){y^2} = 4\left( {x + 1} \right) and y2=8(x+2){y^2} = 8\left( {x + 2} \right) which are perpendicular to each other is
View answer and explanation
Correct option: C
x+3=0x + 3 = 0
The correct answer is C (x+3=0x + 3 = 0) because the locus of the intersection of perpendicular tangents to parabolas is the directrix, and the directrix of the combined parabola system is x = -3.
153 Medium
153 of 211
Application +4 | -1
Locus of mid point of chord of parabola x2=4yx^2=4 y having slope 2
View answer and explanation
Correct option: B
x=4x=4
Option B (x=4) is correct because the locus of the midpoint is found using T=S1 and the given slope condition.
154 Medium
154 of 211
Analyzing +4 | -1
The shortest distance between the line x−y=1 x\mathrm{ }-\mathrm{ }y\mathrm{ } = \mathrm{ }1\, and the curve x2=2y{x^2} = 2y is
View answer and explanation
Correct option: B
122\frac{1}{{2\sqrt 2 }}
Option B (122\frac{1}{{2\sqrt 2 }}) is correct because it is the shortest distance found by finding a tangent to the curve parallel to the line and then calculating the perpendicular distance.
155 Medium
155 of 211
Analyzing +4 | -1
The largest value of aa for which the circle x2+y2=a2x^2+y^2=a^2 falls totally in the interior of the parabola y2=4(x+4)y^2=4(x+4) is
View answer and explanation
Correct option: D
232 \sqrt{3}
Option D (232 \sqrt{{3}}) is correct because it represents the largest radius for the circle to be entirely within the parabola, found by setting the discriminant to zero.
156 Medium
156 of 211
Application +4 | -1
The length of the normal chord to the parabola y2=4xy^2=4 x, which subtends right angle at the vertex is
View answer and explanation
Correct option: A
636 \sqrt{3}
Option A (636 \sqrt{3}) is correct because it represents the length of the normal chord to the parabola that subtends a right angle at the vertex, calculated using the properties of the parabola and normal.
157 Medium
157 of 211
Application +4 | -1
If two tangents drawn from a point PP to the parabola y2=4x{y^2} = 4x be such that the slope of one tangent is double of the other, then PP lies on the curve
View answer and explanation
Correct option: B
9x=2y29x = 2{y^2}
Option D (None of these) is correct because the calculated common tangent does not match any of the provided options after correctly determining the value of m.
158 Medium
158 of 211
Application +4 | -1
The equation of a parabola which passes through the intersection of a straight line x+y=0x + y = 0 and the circle x2+y2+4y=0{x^2} + {y^2} + 4y = 0 is
View answer and explanation
Correct option: C
y2=2x{y^2} = 2x
Option C (y2=2x{y^2} = 2x) is correct because it passes through the intersection points (0,0) and (2,-2) of the given line and circle.
159 Medium
159 of 211
Application +4 | -1
If the focus of parabola (y−k)2=4(x−h)(y-k)^2=4(x-h) always lies between the line x+y=1x+y=1 and x+y=3\mathrm{x}+\mathrm{y}=3 then -
View answer and explanation
Correct option: A
0<h+k<20<\mathrm{h}+\mathrm{k}<2
Option A (0<h+k<20<\mathrm{h}+\mathrm{k}<2) is correct because the focus of the parabola must lie between the given lines, leading to this inequality.
160 Medium
160 of 211
Application +4 | -1
The tangent to the parabola y2=16x{y^2} = 16x, which is perpendicular to a line y−3x−1=0y - 3x - 1 = 0, is
View answer and explanation
Correct option: A
3y+x+36=03y + x + 36 = 0
Option A (3y+x+36=03y + x + 36 = 0) is correct because it represents the tangent to the parabola with a slope that is the negative reciprocal of the given line's slope.
161 Medium
161 of 211
Application +4 | -1
A ray of light moving parallel to the xx-axis gets reflected from parabolic mirror whose equation is (y−3)2=8(x+2)(y-3)^2=8(x+2). After reflection, the ray must pass through
View answer and explanation
Correct option: A
(0,3)(0,3)
Option A ((0,3)) is correct because a ray parallel to the x-axis, after reflection from a parabolic mirror, passes through the focus of the parabola.
162 Medium
162 of 211
Application +4 | -1
The mid-point of the chord intercepted on the line 4x−3y+4=04 x-3 y+4=0 by the parabola y2=8xy^2=8 x, is
View answer and explanation
Correct option: A
(5/4,3)(5 / 4,3)
Option A ((5/4,3)(5 / 4,3)) is correct because the midpoint is found by equating the equation of the chord with the given line and solving for the coordinates of the midpoint.
163 Medium
163 of 211
Application +4 | -1
The angle between the tangents drawn from the point (3,4)\left( {3,4} \right) to the parabola y2−2y+4x=0{y^2} - 2y + 4x = 0 is
View answer and explanation
Correct option: A
tan⁡−1(85/7){\tan ^{ - 1}}\left( {8\sqrt 5 /7} \right)
Option A (tan⁡−1(85/7){\tan ^{ - 1}}\left( {8\sqrt 5 /7} \right)) is correct because it is the angle between the tangents calculated using the properties of the parabola and the tangent equation.
164 Medium
164 of 211
Analyzing +4 | -1
Set of values of ' hh ' for which the number of distinct common normals of (x−2)2=4(y−3)(x-2)^2=4(y-3) and x2+y2−2x−hy−c=0(c>0)x^2+y^2-2 x-h y-c=0(c>0) is 3 , is
View answer and explanation
Correct option: D
(10,∞)(10, \infty)
Option D ((10,∞)(10, \infty)) is correct because the condition for the number of distinct common normals of the given parabola and circle to be 3 leads to this interval for the values of 'h'.
165 Medium
165 of 211
Application +4 | -1
If point (2a, a) lies inside the parabola x2−2x−4y+3=0x^2-2 x-4 y+3=0 then interval of values of aa is
View answer and explanation
Correct option: B
(12,32)\left(\frac{1}{2}, \frac{3}{2}\right)
Option B (12,32)\left(\frac{1}{2}, \frac{3}{2}\right) is correct because substituting the point (2a, a) into the inequality derived from the parabola equation yields the correct interval for a.

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166 Medium
166 of 211
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.
167 Medium
167 of 211
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.
168 Medium
168 of 211
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.
169 Medium
169 of 211
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.
170 Medium
170 of 211
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.
171 Medium
171 of 211
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.
172 Medium
172 of 211
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.
173 Medium
173 of 211
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.
174 Medium
174 of 211
Application +4 | -1
If a focal chord of y2=16xy^2=16 x is a tangent to (x−6)2+y2=2(x-6)^2+y^2=2, then the values of the slope of this chord are
View answer and explanation
Correct option: A
{−1,1}\{-1,1\}
Option A ({−1,1}\left\{-1,1\right\}) is correct because the tangents to the circle that pass through the focus of the parabola have slopes of ±1\pm 1.
175 Medium
175 of 211
Application +4 | -1
The tangents to the parabola y2=4axy^2=4 a x at 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) intersect on its axis, then
View answer and explanation
Correct option: B
t1=−t2t_1=-t_2
Option B (t1=−t2t_1=-t_2) is correct because the point of intersection of the tangents lies on the x-axis, implying that the y-coordinate of the intersection point is zero.
176 Medium
176 of 211
Remembering +4 | -1
The polar of the focus of the parabola is
View answer and explanation
Correct option: C
Directrix
Option C (Directrix) is correct because the polar of the focus of a parabola is its directrix.
177 Medium
177 of 211
Application +4 | -1
The equation of the parabola with (−3,0)(-3,0) as focus, and x+5=0x+5=0 as directrix, is
View answer and explanation
Correct option: C
y2=4(x+4)y^2=4(x+4)
Option C (y2=4(x+4)y^2=4(x+4)) is correct because it represents a parabola opening to the right with vertex at (-4, 0) and focus at (-3, 0).
178 Medium
178 of 211
Application +4 | -1
The equation of the common tangent to the hyperbola 3x2−y2=33{x^2} - {y^2} = 3 and to the parabola y2=8x{y^2} = 8x is
View answer and explanation
Correct option: D
2x+y+1=02x + y + 1 = 0
Option D (2x+y+1=02x + y + 1 = 0) is correct because it represents a common tangent to both the hyperbola and the parabola.
179 Medium
179 of 211
Analyzing +4 | -1
If two different tangents of y2=4x{y^2} = 4x are the normals to x2=4by,{x^2} = 4by, then
View answer and explanation
Correct option: B
∣b∣<122\left| b \right| < \frac{1}{{2\sqrt 2 }}
Option B (∣b∣<122\left| b \right| < \frac{1}{{2\sqrt 2 }}) is correct because it represents the condition for two different tangents to the parabola y2=4x{y^2} = 4x to be normals to the parabola x2=4by{x^2} = 4by.
180 Medium
180 of 211
Application +4 | -1
If y=mx+cy=m x+\mathrm{c} is the normal at a point on the parabola y2=8x\mathrm{y}^{2}=8 x whose focal distance is 8 units, then ∣c∣|c| is equal to :
View answer and explanation
Correct option: C
10310 \sqrt{3}
Option C (10310 \sqrt{3}) is correct because it represents the absolute value of c in the normal equation, derived from the given focal distance.
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