Self Studies

Tangents and its Equations Test 47

Result Self Studies

Tangents and its Equations Test 47
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The line y=mx+1y = mx + 1 is a tangent to the curve y2=4xy^{2} = 4x if the value of m is .......
    Solution
    Given, y2=4x2ydydx=4dydx=2yy^{2} = 4x \Rightarrow 2y \dfrac{dy}{dx} = 4 \Rightarrow \dfrac{dy}{dx} = \dfrac{2}{y}
    but slope of tangent is m=2yy=2mm = \dfrac{2}{y} \Rightarrow y = \dfrac{2}{m}
    When y=2m,x=y24=4/m/42=1m2 y= \dfrac{2}{m}, x = \dfrac{y^{2}}{4} = 4/m^{2}_{/4} = \dfrac{1}{m^{2}}
    x=1m2,y=2mx = \dfrac{1}{m^{2}}, y = \dfrac{2}{m} lie on the line 
    y=mx+1y = mx + 1
    2m=m×1m2+12m=1m+1\dfrac{2}{m} = m \times \dfrac{1}{m^{2}} + 1 \Rightarrow \dfrac{2}{m} = \dfrac{1}{m} + 1
    2m=1m=1\dfrac{2}{m} = 1 \Rightarrow m = 1
    Correct answer (A).
  • Question 2
    1 / -0
    The slope of the normal to the curve y=2x2+3sinx y = 2x ^{2} + 3 \sin x at x=0 x = 0 is 
    Solution
    Given, y=2x2+3sinx y = 2x ^{2} + 3 \sin x
    slope =dydx=4x+3cosx = \dfrac{dy}{dx} = 4x + 3 \cos x
    dydyx=0=4(0)=3\dfrac{dy}{dy}|_{x=0} = 4 ( 0 ) = 3  
    slope of normal = 1/dydx=1/3 -1/ \dfrac{dy}{dx} = -1/3
  • Question 3
    1 / -0
    The line y=x+1 y = x + 1 is a tangent to the curve y2=4xy^{2} = 4 x at the point 
    Solution
    Given,
    y2=4x 2ydxdx=4dxdy=2/yy^{2} = 4 x \\ \Rightarrow  2y \dfrac{dx}{dx} = 4 \\ \Rightarrow \dfrac{dx}{dy} = 2/y  
    but given that y=x+1 y = x + 1 is a tangent to the curve its slope = 1(y=mx+c) 1 ( y = mx +c )
    2y=1    y=2\dfrac 2y =1\implies y=2
    x=21=1x=2-1=1
    The line y=x+1 y = x + 1 is a tangent to the curve y2=4xy^{2} = 4 x at the point (1,2)(1,2).
  • Question 4
    1 / -0
    The points on the curve 9y2=x39 y^{2} = x^{3}, where the normal to the curve makes equal intercepts with the axes are ...........
  • Question 5
    1 / -0
    The coordinates of a point P(x, y) lying in the first quadrant of the ellipse x28+y218=1\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{18}=1 so that the area of the triangle formed by the tangent at P and the axes is the smallest are
    Solution
    Let point be (22sinθ,32cosθ)(2\sqrt 2 sin\theta, 3\sqrt 2 cos\theta)
    y=3222tanθy'=\dfrac {-3\sqrt 2}{2\sqrt 2}tan\theta
    y=32tanθy'=\dfrac {-3}{2}tan\theta
    (y32cosθ)=32tanθ(x22sinθ)(y-3\sqrt 2cos\theta)=\dfrac {-3}{2}tan\theta (x-2\sqrt 2 sin\theta)
    x=2sinθx=\dfrac {2}{sin\theta} (x intercept)
    y=3cosθy=\dfrac {3}{cos\theta} (y intercept)
    so area =12×2×3sinθcosθ=\dfrac {1}{2}\dfrac {\times 2\times 3}{sin\theta cos\theta}
    area is min when sin2θ=1sin 2\theta=1
    θ=π4\theta=\dfrac {\pi}{4}
    so point (2,3)

  • Question 6
    1 / -0

    Directions For Questions

    x2a2+y2b2=1\displaystyle \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 is the equation ot an ellipse. Tangents are drawn to the ellipse and it's auxillary circle at the points where a common ordinate cuts them.

    ...view full instructions

    The greatest inclination between the tangents is
    Solution
    The slope of the tangent to the ellipse =bacotθ=-\displaystyle \dfrac{\mathrm{b}}{\mathrm{a}}\cot\theta
     The slope of tangent to the circle is cotθ-\cot\theta
    \therefore the angle ϕ\phi between the tangents is given by
    tanϕ=(ba+1)cotθ1+bacot2θ=(ab)atanθ+bcotθ=ab(atanθbcotθ)2+2ab\tan\phi=\dfrac{(-\dfrac{\mathrm{b}}{\mathrm{a}}+1)\cot\theta}{1+\dfrac{\mathrm{b}}{\mathrm{a}}\cot^{2}\theta}=\dfrac{(\mathrm{a}-\mathrm{b})}{\mathrm{a}\tan\theta+\mathrm{b}\cot\theta}=\dfrac{\mathrm{a}-\mathrm{b}}{(\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta})^{2}+2\sqrt{\mathrm{a}\mathrm{b}}}
    This is maximum when atanθbcotθ=0\sqrt{\mathrm{a}\tan\theta}-\sqrt{\mathrm{b}\cot\theta}=0
    \therefore the maximum angle =tan1(ab2ab)=\displaystyle \tan^{-1}(\dfrac{\mathrm{a}-\mathrm{b}}{2\sqrt{\mathrm{a}\mathrm{b}}})
  • Question 7
    1 / -0
    A curve passes through (2,0)(2, 0) and the slope of the tangent at any point (x,y)(x, y) is x22xx^2 -2x for all values of xx. The point of minimum ordinate on the curve where x>0x > 0 is (a,b)(a, b)'
    Then find the value of a+6ba + 6b.
    Solution
    dydx=x22x\dfrac {dy}{dx} = x^2 - 2x
    y=x33x2+Cy = \dfrac {x^3}{3} - x^2 + C
    Passing through (2, 0)
    C=43\Rightarrow C = \dfrac {4}{3}
    y=x33x2+43\therefore y = \dfrac {x^3}{3} - x^2 + \dfrac {4}{3}
    y=x22x=x(x2)y' = x^2 - 2x = x(x - 2)
    For maxima or minima, y=0y'=0
    x=0,2 \Rightarrow x= 0,2
    y>0y''>0 at x=2x=2
    At x = 2, y takes the minimum value.
    \therefore  minimum value of y is =834+43=0= \dfrac {8}{3} - 4 +\dfrac {4}{3} = 0
    a=2,b=0\therefore a = 2, b = 0
    a+6b=2a + 6b = 2
  • Question 8
    1 / -0
    The value of xx at which tangent to the curve y=x36x2+9x+4, 0x 5y=x^3-6x^2+9x+4,   0\leq x \leq 5 has maximum slope is
    Solution
    y=x36x2+9x+4y=x^3-6x^2+9x+4
    dydx=3x212x+9\dfrac{dy}{dx}=3x^2-12x+9

    Slope of tangent, 
    m=f(x)=3x212x+9m=f(x)=3x^2-12x+9

    For maxima or minima, 
    f(x)=0f'(x)= 0
    6x12=0\Rightarrow 6x-12=0
     x=2\Rightarrow  x=2.

    Now, 
    m(0)=9, m(2)=3m(0)=9,  m(2)=-3 and m(5)=24m(5)=24
    Hence,
    The maximum value is attained at x=5x = 5.

    Hence, option D.
  • Question 9
    1 / -0
    A function y=f(x)y=f(x) has a second order derivative f(x)=6(x1)f''(x)=6(x-1) .
    If its graph passes through the point (2,1)(2,1) and at that point the tangent to the graph is y=3x5y=3x-5, then the function is
    Solution
    f(x)=6(x1)f''(x)=6(x-1)
    f(x)=3x26x+k \Rightarrow f'(x) = 3x^2 -6x+k...(1)
    Slope of tangent at (2,1)(2,1) is 33.
    3=3×226×2+k \Rightarrow 3=3 \times 2^2 -6 \times 2 +k
    k=3 \Rightarrow k=3...(2)
    From (1) and (2)
    f(x)=x33x2+3x+cf(x)=x^3-3x^2+3x+c
    Using f(x)=y=1f(x)=y=1 when x=2x=2, we get
    1=2+cc=11=2+c \Rightarrow c=-1
    Hence, the function is 
    f(x)=x33x2+3x1f(x)=x^3-3x^2+3x-1
    f(x)=(x1)3 \Rightarrow f(x)=(x-1)^3
  • Question 10
    1 / -0
    Suppose a,b,ca,b,c are such that the curve y=ax2+bx+cy = ax^2 + bx + c is tangent to y=3x3y = 3x -3 at (1,0)(1, 0) and is also tangent to y=x+1y = x + 1 at (3,4)(3, 4) then the value of (2ab4c)(2a -b -4c) equals
    Solution
    y=ax2+bx+cy = ax^2 + bx + c
    Since (1,0)(1,0) lies on the curve
     a+b+c=0 \Rightarrow a + b + c = 0 ...... (i)
    Slope of curve =dydx=2ax+b\displaystyle =\frac {dy}{dx} = 2ax + b
    Slope of curve at (1,0)=2a+b=2a+b
    Slope of curve at (3,4)=6a+b=6a+b
    Slope of tangent y=3x3y=3x-3 is 3
    Slope of tangent y=x+1y=x+1 is 1
    (dydx)x=1=3(\dfrac {dy}{dx})_{x=1} = 3 and (dydx)x=3=1(\dfrac {dy}{dx})_{x=3} = 1
    2a+b=32a + b = 3 .... (ii)
    6a+b=16a + b = 1 .... (iii)
    on solving we get a =12,b=4,c=72= -\dfrac {1}{2}, b = 4, c = -\dfrac {7}{2}
    2ab4c=14+14=9\therefore 2a - b - 4c = -1 -4 + 14 = 9
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now