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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Mixed difficulty

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Showing 151-180 of 281 questions

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151 Medium
151 of 281
Understanding +4 | -1
The product of the lengths of subtangent and subnormal at any point of a curve is
View answer and explanation
Correct option: B
Square of the ordinate
Option B (Square of the ordinate) is correct because the product of the lengths of the subtangent and subnormal is equal to y2, where y is the ordinate.
152 Medium
152 of 281
Application +4 | -1
If the tangent to the curve y=x+sin⁡yy = x + \sin y at a point (a,b)\left( {a,b} \right) is parallel to the line joining (0,32)\left( {0,\frac{3}{2}} \right) and (12,2),\left( {\frac{1}{2},2} \right), then
View answer and explanation
Correct option: A
∣b−a∣=1\left| {b - a} \right| = 1
The correct answer is A (∣b−a∣=1\left| {b - a} \right| = 1) because it is found by using the parallel line condition and the curve equation to find the relationship between a and b.
153 Easy
153 of 281
Application +4 | -1
The equation of the tangent to the curve y=1−ex/2y=1-e^{x / 2} at the point of intersection with the yy-axis, is
View answer and explanation
Correct option: A
x+2y=0x+2 y=0
Option A (x+2y=0x+2y=0) is correct because it represents the equation of the tangent line to the curve at the point (0,0) with a slope of -1/2.
154 Easy
154 of 281
Application +4 | -1
The equation of the tangent to the curve y=e−∣x∣y = {e^{ - \left| x \right|}} at the point where the curve cuts the line x=1x = 1 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated equation of the tangent does not match any of the given options.
155 Easy
155 of 281
Analyzing +4 | -1
If the slope of the tangent to the curve y=excos⁡xy = {e^x}\cos x is minimum at x=a,  0≤a≤2  πx = a,\;0 \le a \le 2\;\pi, then the value of aa is
View answer and explanation
Correct option: B
π\pi
Option B (π\pi) is correct because the second derivative of the slope is positive at x=πx = \pi, indicating a minimum slope.
156 Easy
156 of 281
Application +4 | -1
The length of tangent, subtangent, normal and subnormal for the curve y=x2+x−1y = {x^2} + x - 1 at (1, 1) are A,  B,  C and DA,\;B,\;C\,\mathrm{and}\,D respectively, then their increasing order is
View answer and explanation
Correct option: D
B,  A,  D,  CB,\;A,\;D,\;C
Option B (2 units) is correct because it is the length of the side when the rate of change of area equals the rate of change of perimeter.
157 Easy
157 of 281
Application +4 | -1
Equation of a common tangent (if it exists) to x2+y2=2x^2+y^2=2 and x3y3=1x^3 y^3=1 is
View answer and explanation
Correct option: C
x+y=2x+y=2
Option C (x+y=2x+y=2) is correct because it represents a common tangent to both the circle and the curve, found by equating the derivatives and solving for intersection points.
158 Easy
158 of 281
Application +4 | -1
If STST and SNSN are the lengths of the subtangent and the subnormal at the point θ=π2\theta = \frac{\pi }{2} on the curve x=a(θ+sin⁡θ),  y=a(1−cos⁡θ),  a≠1x = a\left( {\theta + \sin \theta } \right),\;y = a\left( {1 - \cos \theta } \right),\;a \ne 1, then
View answer and explanation
Correct option: A
ST=SNST = SN
Option A (ST=SNST = SN) is correct because at θ=π2\theta = \frac{\pi}{2}, the lengths of the subtangent and subnormal are both equal to aa.
159 Easy
159 of 281
Application +4 | -1
The equation of tangent at (−4,−4)(-4,-4) on the curve x2=−4yx^2=-4 y is
View answer and explanation
Correct option: D
2x−y+4=02 x-y+4=0
Option D (2x−y+4=02 x-y+4=0) is the correct answer because it represents the equation of the tangent line to the curve at the given point.
160 Easy
160 of 281
Application +4 | -1
The slope of the tangent to the curve y=∫0xdx1+x3\mathrm{y}=\int_0^{\mathrm{x}} \frac{\mathrm{dx}}{1+\mathrm{x}^3} at the point where x=1\mathrm{x}=1 is -
View answer and explanation
Correct option: A
1/21 / 2
Option A (1/2) is correct because the derivative of the integral gives 11+x3\frac{{1}}{{1+x^3}}, and evaluating at x=1 yields 1/2.
161 Easy
161 of 281
Application +4 | -1
The equation of normal to the curve x2y=x2−3x+6{x^2}y = {x^2} - 3x + 6 at the point with abscissa x=3x = 3 is
View answer and explanation
Correct option: B
27x−3y=7927x - 3y = 79
Option B (27x−3y=7927x - 3y = 79) is correct because it represents the equation of the normal to the curve at the specified point, calculated using the derivative and the negative reciprocal slope.
162 Easy
162 of 281
Analyzing +4 | -1
The intercept on xx-axis made by tangents to the curve y=∫0x∣t∣dt,x∈Ry=\int\limits_{0}^x|t| d t, x \in R, which are parallel to the line y=2xy=2 x, are equal to
View answer and explanation
Correct option: A
±1\pm 1
Option A (±1\pm 1) is correct because the x-intercepts of the tangents parallel to y=2x are ±1\pm 1.
163 Medium
163 of 281
Application +4 | -1
The slope of the tangent to the curve y=9−x2y = \sqrt {9 - {x^2}} at the point where ordinate and abscissa are equal, is
View answer and explanation
Correct option: B
−1- 1
Option B (-1) is correct because it represents the slope of the tangent line at the point where the x and y coordinates are equal on the given curve.
164 Medium
164 of 281
Application +4 | -1
The angle of intersection of the curves y=x2,  y=7−x3y = {x^2},\;y = 7 - {x^3} at (1, 1) is
View answer and explanation
Correct option: C
π/2\pi /2
Option C (π/2\pi /2) is correct because the product of the slopes of the tangents to the two curves at the point of intersection is -1, indicating perpendicularity.
165 Hard
165 of 281
Application +4 | -1
The equation of a curve is y=f(x)y=f(x). The tangents at (1,f(1)),(2,f(2))(1, f(1)),(2, f(2)) and (3,f(3))(3, f(3)) make angles π6,π3\frac{\pi}{6}, \frac{\pi}{3} and π4\frac{\pi}{4} respectively with the positive direction of the xx-axis. Then, the value of ∫23f′(x)f′′(x)dx+∫13f′′(x)dx\int\limits_{2}^3 f^{\prime}(x) f^{\prime \prime}(x) d x+\int\limits_{1}^3 f^{\prime \prime}(x) d x is equal to
View answer and explanation
Correct option: A
−13-\frac{1}{\sqrt{3}}
Option A (−13-\frac{1}{\sqrt{3}}) is correct because it is the result of evaluating the definite integral using the given derivative values.

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166 Easy
166 of 281
Application +4 | -1
If an equation of a tangent to the curve, y−cos⁡(x+y),−1−1≤x≤1+πy-\cos (x+y),-1-1 \leq x \leq 1+\pi, is x+2y=kx+2 y=k then kk is equal to :
View answer and explanation
Correct option: D
π2\frac{\pi}{2}
Option D (π2\frac{{\pi}}{{2}}) is correct because it is the value of k obtained by finding the point on the curve where the tangent has the given slope and substituting it into the tangent equation.
167 Medium
167 of 281
Application +4 | -1
If the line xcos⁡α+ysin⁡α=Px \cos \alpha+y \sin \alpha=P touches the curve 4x3=27ay24 x^{3}=27 a y^{2}, then Pa=\frac{P}{a}= :
View answer and explanation
Correct option: A
cot⁡2αcos⁡α\cot ^{2} \alpha \cos \alpha
Option A (cot⁡2αcos⁡α\cot ^{2} \alpha \cos \alpha) is correct because it is derived by finding the point of tangency and substituting it into the line equation.
168 Medium
168 of 281
Application +4 | -1
The angle between the tangents to the curve y2=2ax{y^2} = 2ax at the point where x=a2x = \frac{a}{2}, is
View answer and explanation
Correct option: D
π2\frac{\pi }{2}
Option D (π2\frac{\pi }{2}) is correct because the product of the slopes of the tangents at the given points is -1, indicating perpendicularity.
169 Medium
169 of 281
Analyzing +4 | -1
Tangents are drawn from the origin to the curve y=cos⁡  xy = \cos \;x. Their points of contact lie on
View answer and explanation
Correct option: C
x2y2=x2−y2{x^2}{y^2} = {x^2} - {y^2}
The correct answer is C (x2y2=x2−y2{x^2}{y^2} = {x^2} - {y^2}) because it represents the locus of the points of contact of tangents drawn from the origin to the curve y=cos⁡xy = \cos x.
170 Easy
170 of 281
Analyzing +4 | -1
The angle between the curves xy=2x y=2 and y2=4xy^2=4 x at their point of intersection is
View answer and explanation
Correct option: B
tan⁡−13\tan ^{-1} 3
The correct answer is B (tan⁡−13\tan^{-1} 3) because by finding the slopes of the tangents at the point of intersection and using the formula for the angle between two lines, the angle is tan⁡−13\tan^{-1} 3.
171 Easy
171 of 281
Application +4 | -1
The points at which the tangents to the curve ax2+a x^2+ 2hxy+by2=12 h x y+b y^2=1 is parallel to yy-axis is
View answer and explanation
Correct option: B
where hx+by=0h x+b y=0 meets it
Option B (where hx+by=0hx+by=0 meets it) is correct because the tangent is parallel to the y-axis when the denominator of the derivative is zero.
172 Easy
172 of 281
Application +4 | -1
If the tangent at any point on the curve x4+y4=c4x^4+y^4=c^4 cuts off intercepts aa and bb on the coordinate axes, the value of a−4/3+b−4/3a^{-4 / 3}+b^{-4 / 3} is
View answer and explanation
Correct option: A
c−4/3c^{-4 / 3}
Option A (c−4/3c^{-4 / 3}) is correct because the equation of the tangent and the given curve are used to find the intercepts and simplify the expression.
173 Hard
173 of 281
Application +4 | -1
The sum of intercepts on the coordinate axes made by the tangent to the curve x+y=a\sqrt{x}+\sqrt{y}=\sqrt{a} is
View answer and explanation
Correct option: A
aa
Option A (aa) is correct because the sum of the intercepts of the tangent to the curve is equal to aa, as derived from the given equation.
174 Medium
174 of 281
Application +4 | -1
The tangent line at (−1,15)(-1,15) to the curve y=2x3+13x2+5x+9y=2 x^{3}+13 x^{2}+5 x+9 passes through
View answer and explanation
Correct option: A
Origin
Option A (Origin) is correct because the equation of the tangent line at (-1, 15) is y + 15x = 0, which passes through the origin (0,0).
175 Easy
175 of 281
Application +4 | -1
The length of the normal to the curve x=a(θ+sin⁡θ),y=a(1−cos⁡θ)x=a(\theta+\sin \theta), y=a(1-\cos \theta), at θ=π/2\theta=\pi / 2, is
View answer and explanation
Correct option: B
a2a \sqrt{2}
Option B (a2a \sqrt{2}) is correct because the length of the normal to the curve at θ=π/2\theta = \pi/2 is calculated to be a2a\sqrt{2}.
176 Medium
176 of 281
Application +4 | -1
The angle between curves y2=4xy^2=4 x and x2+y2=5x^2+y^2=5 at (1,2)(1,2) is
View answer and explanation
Correct option: A
tan⁡−1(3)\tan ^{-1}(3)
Option A (tan⁡−1(3)\tan^{-1}(3)) is correct because by finding the derivatives of both curves at the point (1,2) and using the formula for the angle between two lines, we can find the angle between the curves.
177 Easy
177 of 281
Analyzing +4 | -1
Any tangent to the curve y=2x5+4x3+7x+9y = 2{x^5} + 4{x^3} + 7x + 9
View answer and explanation
Correct option: C
Makes an acute angle with the xx-axis
Option A (Parallel) is correct because the tangents at the points where the curve crosses the x-axis have the same slope.
178 Medium
178 of 281
Application +4 | -1
If the tangent to the curve y=f(x)=xlog⁡ex,(x>0)y = f\left( x \right) = x{\log _e}x,\left( {x > 0} \right) at a point (c,f(c))\left( {c,f\left( c \right)} \right) is parallel to the line segment joining to the points(1,0) and (e, e) then c=
View answer and explanation
Correct option: B
e(1e−1){e^{\left( {\frac{1}{{e - 1}}} \right)}}
Option B (e^(1/(e-1))) is correct because the derivative of the function at point c must equal the slope of the line segment joining the two points, which can be solved for c.
179 Easy
179 of 281
Application +4 | -1
For the curve x=t2−1,y=t2−tx=t^2-1, y=t^2-t, the tangent line is perpendicular to xx-axis when
View answer and explanation
Correct option: A
t=0t=0
Option A (t=0) is correct because the tangent line is perpendicular to the x-axis when dx/dt = 0 and dy/dt is not 0.
180 Medium
180 of 281
Analyzing +4 | -1
If the tangent at each point of the curve y=23x3−2ax2+2x+5y=\frac{2}{3} x^3-2 a x^2+2 x+5 makes an acute angle with the positive direction of xx-axis, then
View answer and explanation
Correct option: B
−1≤a≤1-1 \leq a \leq 1
Option B (−1≤a≤1-1 \leq a \leq 1) is correct because the derivative of the curve must be non-negative for all x, leading to the given range for 'a'.
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