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Tangents and its Equations Test 51

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Tangents and its Equations Test 51
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  • Question 1
    1 / -0
    If the sum of the squares of the intercepts on the axes cut off by the tangent to the cuve x1/3+y1/3=a1/3(a>0)\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right ) at (a8,a8)\displaystyle \left ( \dfrac{a}{8}, \dfrac{a}{8} \right ) is 22, then aa has the value
    Solution
    x1/3+y1/3=a1/3(a>0)\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )
    y(x)=y2/3x2/3y'(x)= -\cfrac{y^{2/3}}{x^{2/3}}
    y(x,y)=(a8,a8)=1\left.y'\right|_{(x,y)=\left(\dfrac{a}{8}, \dfrac{a}{8}\right)} = -1

    Equation of tangent
    x+y=c x+y =c
    x+y=a4;      (a8,a8)x +y = \dfrac{a}{4};\>\>\>\>\>\> \because \left(\dfrac{a}{8}, \dfrac{a}{8}\right) satisfies it

    xx-intercept =aa4=y= a\dfrac{a}{4} = y-intercept
    Given, a216+a216=2 \cfrac{a^2}{16}+\cfrac{a^2}{16} = 2
    a= ±4a =  \pm 4
    a=4;     a>0a =4;\>\>\>\>\>\because a>0
  • Question 2
    1 / -0
    The coordinates of the point PP on the curve y2=2x3y^{2}= 2x^{3} the tangent at which is perpendicular to the line 4x3y+2=04x-3y + 2 = 0, are given by
    Solution
    4x3y+2=04x-3y+2=0
    Or 
    3y=4x+23y=4x+2
    Or 
    y=43x+23y=\dfrac{4}{3}x+\dfrac{2}{3}.
    Hence slope of the tangent will be 
    =34=\dfrac{-3}{4}
    Now 
    dydx=34\dfrac{dy}{dx}=\dfrac{-3}{4}
    y=±2.x32y=\pm\sqrt{2}.x^{\frac{3}{2}}
    Hence
    dydx\dfrac{dy}{dx}

    =±2.32.x12=\pm\sqrt{2}.\dfrac{3}{2}.x^{\frac{1}{2}}

    =34=\dfrac{-3}{4}

    Or 
    2.32.x12=34-\sqrt{2}.\dfrac{3}{2}.x^{\dfrac{1}{2}}=\dfrac{-3}{4}
    Or 
    22.x12=12\sqrt{2}.x^{\dfrac{1}{2}}=1
    Or 
    x=18x=\dfrac{1}{8}.
    Hence
    y=2.x32y=-\sqrt{2}.x^{\dfrac{3}{2}}
    =2.1162=-\sqrt{2}.\dfrac{1}{16\sqrt{2}}
    =116=-\dfrac{1}{16}
    Hence the co-ordinates are  (18,116)(\dfrac{1}{8},\dfrac{-1}{16})
  • Question 3
    1 / -0
    The tangent to the curve x=acos2θcosθx= a\sqrt{\cos 2\theta }\cos \theta , $$y= a\sqrt{\cos 2\theta }\sin \theta
    atthepointcorrespondingto at the point corresponding to \theta = \pi /6$$ is
    Solution
    dxdθ=acos2θsinθ +acosθsin2θcos2θ\displaystyle \dfrac{dx}{d\theta }= -a\sqrt{\cos 2\theta }\sin \theta  +\dfrac{-a\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}

    =acos2θsinθ+cosθsin2θcos2θ\displaystyle = -a\dfrac{\cos 2\theta \sin \theta +\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}

    =asin3θcos2θ\displaystyle = \dfrac{-a\sin 3\theta }{\sqrt{\cos 2\theta }}

    dydθ=acos2θcosθasinθsin2θcos2θ\displaystyle \dfrac{dy}{d\theta }= a\sqrt{\cos 2\theta }\, \cos \theta -a\dfrac{\sin \theta \sin 2\theta }{\sqrt{\cos 2\theta }}

    $$
    \displaystyle = \dfrac{a\cos 3\theta }{\sqrt{\cos 2\theta }}$$

    Hence dydx=cot3θdydxθ=π/6=0\displaystyle \dfrac{dy}{dx}= -\cot 3\theta \Rightarrow \dfrac{dy}{dx}|_{\theta = \pi /6 }= 0
    So the tangent to the curve at θ=π/6\theta = \pi /6 is parallel to the xx-axis.
  • Question 4
    1 / -0
    The equation of the tangent line at an inflection point of f(x)=x46x3+12x28x+3\displaystyle f\left ( x \right )=x^{4}-6x^{3}+12x^{2}-8x+3 is
    Solution
    Given, f(x)=x46x3+12x28x+3\displaystyle f\left ( x \right )=x^{4}-6x^{3}+12x^{2}-8x+3
    At inflection point f(x)=0 f''(x) = 0
    12x236x+24=0 12x^2-36x+24 = 0
    (x2)(x1)=0(x-2)(x-1) = 0
    x=2,1x =2,1
    f(x) =4x318x2+24x8f'(x)  = 4x^3-18x^2+24x-8
    f(1)=2f'(1) = 2 and f(2)=0f'(2) = 0
    f(1)=2f(1) = 2 and f(2)=3 f(2) = 3 
    So the equations are y=3 y=3 and y=2x y=2x
  • Question 5
    1 / -0
    The equation of the common tangent to the curves y2=8x\displaystyle y^{2}= 8x and xy=1 \displaystyle xy= -1 is
    Solution
    A point on the curve xy=1xy = -1 is of the form (t,1/t)\left ( t,-1/t \right ).Now

    y=1xdydx=1x2dydx( t,1/t)=1t2\displaystyle y= -\dfrac{1}{x}\Rightarrow \dfrac{dy}{dx}= \dfrac{1}{x^{2}}\Rightarrow \dfrac{dy}{dx}|_{\left (  t,-1/t \right )}= \dfrac{1}{t^{2}}

    \therefore Equation of tangent to the curve xy=1xy = -1 at  ( t,1/t)\left (  t,-1/t \right ) is

    y+1t=1t2(xt)y=1t2x1t\displaystyle y+\dfrac{1}{t}= \dfrac{1}{t^{2}}\left ( x-t \right )\Rightarrow y= \dfrac{1}{t^{2}}\, x-\dfrac{1}{t}

    For this line to be tangent to the parabola y2=8xy^{2}= 8x it should
    be of the form y=mx+2my=mx+\displaystyle \dfrac{2}{m} so 

    1t2\displaystyle \dfrac{1}{t^{2}} and 2t=2m\displaystyle \dfrac{-2}{t}= \dfrac{2}{m} m=1m=-1

    t=1t2t3=1t=1\displaystyle \therefore -t= \dfrac{1}{t^{2}}\Rightarrow -t^{3}= 1\Rightarrow t= -1

    Thus the required tangent is y=x+2y = x + 2.
  • Question 6
    1 / -0
    The point(s) on the curve y3+3x2=12yy^{3}+3x^{2}=12y the tangent is vertical is (are)
    Solution
    3y2.y+6x=12y3y^{2}.y'+6x=12y'
    Or 
    y(3y212)=6xy'(3y^{2}-12)=-6x
    Or 
    y(y24)=2xy'(y^{2}-4)=-2x
    Or 
    y=2xy24y'=\dfrac{-2x}{y^{2}-4}
    Hence for the tangent to be vertical
    y24=0y^{2}-4=0
    y=±2y=\pm2
    Substituting y=2y=2, we get 
    3x2=12yy33x^{2}=12y-y^{3}
    =248=24-8
    =16=16
    Hence
    x=±43x=\dfrac{\pm4}{\sqrt{3}}
    y=2y=-2 gives 
    3x2=163x^{2}=-16 
    This is not possible.
    Hence the required point is 
    (±43,2)(\dfrac{\pm4}{\sqrt{3}},2).
  • Question 7
    1 / -0
    The equation of the tangents to 4x29y2=36\displaystyle 4x^{2}-9y^{2}=36 which are perpendicular to the straight line 2y+5x=10\displaystyle 2y+5x= 10 are
    Solution
    Given, 2y+5x=102y+5x=10
    2y+5=02y'+5=0
    y=52y'=\dfrac{-5}{2}.
    Hence the required slope is 25\dfrac{2}{5}.
    Now 
    4x29y2=364x^{2}-9y^{2}=36.
    4x236=9y24x^{2}-36=9y^{2}
    y=13.4x236y=\dfrac{1}{3}.\sqrt{4x^{2}-36}.
    y=8x64x236y'=\dfrac{8x}{6\sqrt{4x^{2}-36}} =25=\dfrac{2}{5}
    40x=124x23640x=12\sqrt{4x^{2}-36}
    10x=34x23610x=3\sqrt{4x^{2}-36}
    100x2=36x2324100x^{2}=36x^{2}-324
    64x2=32464x^{2}=-324
    x2=8116x^{2}=\dfrac{-81}{16}.
    Hence no real values of x.
    Hence answer is none of these.
  • Question 8
    1 / -0
    Of all the line tangent to the graph of the curve  y=6x2+3,\displaystyle y=\frac{6}{x^{2}+3}, find the equations of the tangent lines of minimum and maximum slope respectively.
    Solution
    Slope of the tangents will be 
    y=12x(x2+3)2y'=\dfrac{-12x}{(x^{2}+3)^{2}}
    Now for maximum and minimum slopes
    y"=0y"=0
    Or 
    12(x2+3)2+24x(x2+3).2x=0-12(x^{2}+3)^{2}+24x(x^{2}+3).2x=0
    Or 
    (x2+3)[48x212x236]=0(x^{2}+3)[48x^{2}-12x^{2}-36]=0
    Or 
    36x236=036x^{2}-36=0
    Or 
    x=±1x=\pm1.
    Hence
    y(1)=1216=34y'(1)=\dfrac{-12}{16}=\dfrac{-3}{4}
    And 
    y(1)=34y'(-1)=\dfrac{3}{4}.
    Hence y(1)=y(1)=64=32y(1)=y(-1)=\dfrac{6}{4}=\dfrac{3}{2}.
    Hence the equations are 
    y32=34(x+1)y-\dfrac{3}{2}=\dfrac{3}{4}(x+1)
    Or 
    4y6=3x+34y-6=3x+3
    Or 
    3x4y+9=03x-4y+9=0
    And 
    y32=34(x1)y-\dfrac{3}{2}=\dfrac{-3}{4}(x-1)
    Or 
    4y6=3x+34y-6=-3x+3
    Or 
    3x+4y6=03x+4y-6=0.
  • Question 9
    1 / -0

    Directions For Questions

    Let y=f(x)y = f(x) be a differentiable function which satisfies f(x)=f2(x)\displaystyle f'(x)=f^{2}(x) and f(O)=12\displaystyle f(O)=-\frac{1}{2} The graph of the differentiable function y=g(x)y = g(x) contains the point (0,2)(0, 2) and has the property that for each number 'P' the line tangent to y=g(x)y = g(x) at (P,g(p))(P, g(p)) intersetct x - axis at  P+2P + 2
    On the basis of above information answer the following questions

    ...view full instructions

    If the tangent is drawn to the curve y = f(x) at a point where it crosses the y - axis then its equation is
    Solution
    It is given that 
    df(x)dx=f(x)2\dfrac{df(x)}{dx}=f(x)^{2}
    Let 
    f(x)=yf(x)=y
    Then
    y=y2y'=y^{2}
    Or 
    y2.dy=dx\int y^{-2}.dy=\int dx
    Or 
    1y=x+c\dfrac{-1}{y}=x+c
    Now 
    y=12y=\dfrac{-1}{2} at x=0.
    Hence
    c=2c=2
    Or 
    1y=x+2-\dfrac{1}{y}=x+2
    Or 
    xy+2y+1=0xy+2y+1=0 ...(i)
    dydxx=0\dfrac{dy}{dx}_{x=0}
    =yy=122=y^{2}_{y=\frac{-1}{2}}
    Or 
    y=14y'=\dfrac{1}{4}.
    Now equation of the tangent is 
    y(12)=14(x0)y-(\dfrac{-1}{2})=\dfrac{1}{4}(x-0)
    Or 
    4y+2=x4y+2=x
    Or 
    x4y=2x-4y=2 is the required equation of tangent.
  • Question 10
    1 / -0
    What normal to the curve y=x2\displaystyle y=x^{2} form the shortest chord?
    Solution
    Any point on curve y=x2\displaystyle y=x^{2} is P(t,t2)\displaystyle P\left ( t,t^{2} \right )
    $$\displaystyle

    \frac{dy}{dx}=2xequationofnormalat  equation of normal at \displaystyle \left (

    t,t^{2} \right )is is\displaystyle y-t^{2}=-\frac{1}{2t}\left ( x-t

    \right )$$
    solving with y=x2\displaystyle y=x^{2} we get
    $$\displaystyle

    x^{2}-t^{2}=\frac{-1}{2t}\left ( x-t \right )\Rightarrow \left ( x-t

    \right )\left ( x+t+\frac{1}{2t} \right )=0\Rightarrow

    x=-t-\frac{1}{2t}$$
    So normal cuts the curve again at Q(t12t,(t12t)2)\displaystyle Q\left ( -t-\frac{1}{2t},\left ( -t-\frac{1}{2t} \right )^{2} \right )
    z=PQ2=4t2(1+14t2)3\displaystyle z=PQ^{2}=4t^{2}\left ( 1+\frac{1}{4t^{2}} \right )^{3}
    $$\displaystyle

    \frac{dz}{dt}=0\Rightarrow t=\pm \frac{1}{\sqrt{2}}=0,\frac{dz}{dt}$$

    changes sign from negative to positive about $$\displaystyle

    t=\frac{1}{\sqrt{2}}aswellas  as well as \displaystyle

    t=-\frac{1}{\sqrt{2}}$$(No chord is format for t=0)
    z is minimum

    at t=±12\displaystyle t=\pm \frac{1}{\sqrt{2}} & minimum value

    of z=PQ2=3\displaystyle z=PQ^{2}=3 shortest normal chord has

    length 3\displaystyle \sqrt{3} & its equation is $$\displaystyle

    x+\sqrt{2y}-\sqrt{2}=0or  or \displaystyle x-\sqrt{2y}+\sqrt{2}=0$$
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