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Tangents and its Equations Test 20

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Tangents and its Equations Test 20
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  • Question 1
    1 / -0
    A and B are points (2,0)(-2,0) and (1,3)(1,3) on the curve y=4x2\displaystyle y=4-x^{2}. If the tangent at P on the curve be parallel to chord AB, then co-ordinates of point P are 
    Solution
    Given equation of curve y=4x2\displaystyle y=4-x^{2}     ...(i)
    dydx=2x\dfrac{dy}{dx}=-2x

    Given points A(2,0)A(-2,0) and (1,3)(1,3)
    Slope of AB =33=1=\dfrac{3}{3}=1

    2x=1\Rightarrow -2x=1
    x=12\Rightarrow x=\dfrac{-1}{2}

    So, by equation (i), we get
    y=154y=\dfrac{15}{4}
    Hence, the point is (12,154)(-\dfrac{1}{2},\dfrac{15}{4})
  • Question 2
    1 / -0
    The parametric equations of a curve are given by x=sec2t,y=cott.\displaystyle x= \sec ^{2}t, y= \cot t. Tangent at Pt=π4\displaystyle P \,t=\dfrac { \pi }4 meets the curve again at Q; then PQ=?\displaystyle PQ=?
    Solution
    x=sec2t,y=cott.\displaystyle x= \sec ^{2}t, y= \cot t. 
    Given t=π4t=\dfrac{\pi}{4}
    x=sec2(π4),y=cotπ4\Rightarrow x=\sec^2 (\dfrac{\pi}{4}), y=\cot \dfrac{\pi}{4}
    x=2,y=1\Rightarrow x=2, y=1
    So, the point P is (2,1)(2,1)

    Now, dxdt=2sectsecttant\dfrac{dx}{dt}=2\sec t \sec t\tan t
     dydt=cosec2t\dfrac{dy}{dt}=-cosec^2 t
    dydx=12cot3t\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{2}\cot^3 t
    Slope of tangent at P is 12-\dfrac{1}{2}

    Equation of tangent at P is 
    y1=12(x2)y-1=-\dfrac{1}{2}(x-2)
    x+2y=4\Rightarrow x+2y=4          .....(1)

    Since, sec2ttan2t=1\sec^2 t -\tan^2 t=1
    x1y2=1\Rightarrow x-\dfrac{1}{y^2}=1      .....(2)

    Solving (1) and (2), we get
    (42y)y21=y2.\displaystyle \left ( 4-2y \right )y^{2}-1= y^{2}. 
    y=1,1,12\Rightarrow \displaystyle y= 1, 1,-\frac{1}{2}
    x=2,5\Rightarrow x=2, 5    (by (1))

    Qis(5,12)\displaystyle \therefore Q is\left ( 5,-\frac{1}{2} \right )
    PQ=454=352.\therefore PQ= \sqrt{\dfrac{45}{4}}= \dfrac{3\sqrt{5}}{2}. 
    The values of y=1,1y=1,1 correspond to point P.
  • Question 3
    1 / -0
    The line xa+yb=1\dfrac xa+\dfrac yb=1 touches the curve y=bex/a\displaystyle y=be^{-x/a} at the point
    Solution
    Simplifying the equation of the line we get 
    bx+ay=abbx+ay=ab
    ay=bx+abay=-bx+ab
    Or 
    y=ba.x+by=\dfrac{-b}{a}.x+b
    Hence
    dydx=ba\dfrac{dy}{dx}=\dfrac{-b}{a}
    Or 
    ba.ex/a.=ba-\dfrac{b}{a}.e^{-x/a}.=\dfrac{-b}{a}
    Or 
    ex/a=1e^{-x/a}=1
    Or 
    x=0x=0
    Hence
    y=by=b.
    Therefore the point is (0,b)(0,b)
  • Question 4
    1 / -0
    The curve yexy+x=0\displaystyle y-e^{xy}+x=0 has a vertical tangent at the point 
    Solution
    dydxexy[xdydx+y]+1=0\dfrac{dy}{dx}-e^{xy}[x\dfrac{dy}{dx}+y]+1=0
    Or 
    dydx[1x.exy]+1y.exy=0\dfrac{dy}{dx}[1-x.e^{xy}]+1-y.e^{xy}=0
    Or 
    dydx=(1y.exy)1x.exy\dfrac{dy}{dx}=\dfrac{-(1-y.e^{xy})}{1-x.e^{xy}}
    Now 
    1xexy=01-xe^{xy}=0 since the slope of the tangent is 90090^{0}.
    Hence
    xexy=1xe^{xy}=1
    Or 
    x(x+y)=1x(x+y)=1
    Or 
    x2+xy=1x^{2}+xy=1
    Or 
    y=1x2xy=\dfrac{1-x^{2}}{x}
    Or 
    y=1xxy=\dfrac{1}{x}-x ...(i)
    Considering y=0y=0,
    x2=1x^{2}=1
    x=±1x=\pm1.
    Hence we get 2 points (1,0)(1,0) and (1,0)(-1,0).
    Out of these 2, only (1,0)(1,0) lies on the curve.
    Hence the required point is (1,0)(1,0).
  • Question 5
    1 / -0
    If xcosα+ysinα=p\displaystyle x\cos \alpha +y\sin \alpha =p touches x2+a2y2=a2,\displaystyle x^{2}+a^{2}y^{2}=a^{2}, then
    Solution
    Solving for y, we have 
    y=psinα  xcotα \displaystyle y=\dfrac { p }{ \sin { \alpha  }  } -x\cot { \alpha  }
    Putting this value in x2+a2y2=a2{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }, we have
    x2+a2(psinα  xcotα  ) 2=a2\displaystyle { x }^{ 2 }+{ a }^{ 2 }{ \left( \dfrac { p }{ \sin { \alpha  }  } -x\cot { \alpha  }  \right)  }^{ 2 }={ a }^{ 2 }

    x2(1+a2cot2α  )2a2xpcotα  sinα  +a2p2sin2α  a2=0\displaystyle \Rightarrow { x }^{ 2 }\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha  }  \right) -\dfrac { 2a^{ 2 }xp\cot { \alpha  }  }{ \sin { \alpha  }  } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha  }  } -{ a }^{ 2 }=0

    The discriminant of this equation must be zero. So
    a4p2cot2α  sin2α  =(1+a2cot2α  )(a2p2sin2α  a2)a2p2cot2α =(1+a2cot2α  )(p2sin2α  )p2(a2cot2α 1a2cot2α  )=sin2α a2cos2α p2=a2cos2α +sin2α \displaystyle { a }^{ 4 }\dfrac { { p }^{ 2 }\cot ^{ 2 }{ \alpha  }  }{ \sin ^{ 2 }{ \alpha  }  } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha  }  \right) \left( \dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha  }  } -{ a }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }{ p }^{ 2 }\cot ^{ 2 }{ \alpha  } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha  }  \right) \left( { p }^{ 2 }-\sin ^{ 2 }{ \alpha  }  \right) \\ \Rightarrow { p }^{ 2 }\left( { a }^{ 2 }\cot ^{ 2 }{ \alpha  } -1-{ a }^{ 2 }\cot ^{ 2 }{ \alpha  }  \right) =\sin ^{ 2 }{ \alpha  } -{ a }^{ 2 }\cos ^{ 2 }{ \alpha  } \\ \Rightarrow { p }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \alpha  } +\sin ^{ 2 }{ \alpha  }
  • Question 6
    1 / -0
    The point of intersection of the tangents drawn to the curve x2y=1y\displaystyle x^{2}y=1-y at the points where it is met by the curve xy=1y\displaystyle xy=1-y is given by
    Solution
    Intersection of x2y=1y{ x }^{ 2 }y=1-y are xy=1yxy=1-y
    are given by x2y=xy{ x }^{ 2 }y=xy.
    Since y0y\neq 0 on any of two curves
    So we must have x=0,1x=0,1.
    Thus the curve intersect at (0,1);(0,12 )\displaystyle \left( 0,1 \right) ;\left( 0,\dfrac { 1 }{ 2 }  \right)

    Now differentiating x2y=1y{ x }^{ 2 }y=1-y. we have
    dydx=2xy1+x2\displaystyle \dfrac { dy }{ dx } =\dfrac { 2xy }{ 1+{ x }^{ 2 } }
    Thus for (x,y)=(0,1)\left( x,y \right) =\left( 0,1 \right) dydx=0\displaystyle \dfrac { dy }{ dx } =0
    And for (x,y)=(0,12 )\displaystyle \left( x,y \right) =\left( 0,\dfrac { 1 }{ 2 }  \right) dydx=12\displaystyle \dfrac { dy }{ dx } =-\dfrac { 1 }{ 2 }
    Thus the equation of tangent are Y=1Y=1 and Y12=(12 )(X1)\displaystyle Y-\dfrac { 1 }{ 2 } =\left( -\dfrac { 1 }{ 2 }  \right) \left( X-1 \right)
    Their intersection is given by (0,1)\left( 0,1 \right)
  • Question 7
    1 / -0
    The coordinates of the point M(x,y)M(x, y) on y=ex\displaystyle y= e^{-\left | x \right |} so that the area formed by the coordinates axes and the tangent at MM is greatest, are
    Solution
    The curve of f(x) is symmetrical about y axis.
    Let the point of contact of the tangent be (a,ea)(a,e^{-a})
    Therefore
    dydx(a,ea)=ea\dfrac{dy}{dx}_{(a,e^{-a})}=-e^{-a}
    Hence the equation of the tangent will be 
    yea=ea(xa)y-e^{-a}=-e^{-a}(x-a)

    yea=eax+aeay-e^{-a}=-e^{-a}x+ae^{-a}

    y+eax=ea(1+a)y+e^{-a}x=e^{-a}(1+a)

    yea(1+a)+x1+a=1\dfrac{y}{e^{-a}(1+a)}+\dfrac{x}{1+a}=1
    Hence the area formed by the tangent and the coordinate axes will be
    =(xintercept)(yintercept)2=\dfrac{(x-intercept)(y-intercept)}{2}

    A=ea(1+a)22A=\dfrac{e^{-a}(1+a)^{2}}{2}

    Differentiating with respect to a, we get
    dAda=12[2(1+a)ea(1+a)2ea]\dfrac{dA}{da}=\dfrac{1}{2}[2(1+a)e^{-a}-(1+a)^{2}e^{-a}]

    =(1+a)ea2[2(1+a)]=\dfrac{(1+a)e^{-a}}{2}[2-(1+a)]

    =(1+a)ea2(1a)=\dfrac{(1+a)e^{-a}}{2}(1-a)
    =0=0
    Hence
    a=±1a=\pm1
    Thus we get the points as (1,e1)(1,e^{-1}) and (1,e1)(-1,e^{-1}).
  • Question 8
    1 / -0
    The value of m for which the area of the triangle included between the axes and any tangent to the xmy=bm\displaystyle x^{m}y= b^{m} curve is constant, is
    Solution
    Given, y=bmxmy =b^mx^{-m}
    y=bmx(m+1)y' = -b^mx^{-(m+1)}
    At any point (x1,y1)(x1,y1)
    Equation of tangent is given by
    yy1xx1=bmx1(m+1)\cfrac{y-y1}{x-x1} = -b^mx1^{-(m+1)}
    Y intercept =m= -m
    X intercept =(m1)×1m= (m-1)\times \dfrac 1m
    Area bounded = 12(m1)x1m×m=  \cfrac{1}{2}\cfrac{(m-1)x1}{m} \times m
    For area to be constant, m=1m =1
  • Question 9
    1 / -0
    If the tangent at any point on the curve x4+y4=a4\displaystyle x^{4}+y^{4}= a^{4} cuts off intercepts p and q on the coordinate axes, the value of p43+q43\displaystyle p^{-\tfrac 43}+q^{-\tfrac 43} is
    Solution

    Let 
    x=acosθx=a\sqrt{\cos \theta}
    y=asinθy=a\sqrt{\sin \theta}
    Hence
    dydx\dfrac{dy}{dx}

    =cosθ.cosθsinθ.sinθ=-\dfrac{\cos \theta.\sqrt{\cos \theta}}{\sin \theta.\sqrt{\sin \theta}}
    =cotθ32=\cot\theta^{\dfrac{3}{2}}.
    Hence equation of tangent will be 
    yasinθ=cosθ.cosθsinθ.sinθ(xacosθ)y-a\sqrt{\sin \theta}=-\dfrac{\cos \theta.\sqrt{\cos \theta}}{\sin \theta.\sqrt{\sin \theta}}(x-a\sqrt{\cos \theta})
    sinθ.sinθyasin2θ=cosθcosθx+acos2θ\sin \theta.\sqrt{\sin \theta}y-a\sin ^{2}\theta=-\cos \theta\sqrt{\cos \theta}x+a\cos ^{2}\theta
    (sinθ)32.y+(cosθ)32x=a(\sin \theta)^{\dfrac{3}{2}}.y+(\cos \theta)^{\dfrac{3}{2}}x=a
    Hence the intercepts are 
    Q=a(sinθ)32Q=\dfrac{a}{(\sin \theta)^{\dfrac{3}{2}}} and P=a(cosθ)32P=\dfrac{a}{(\cos \theta)^{\dfrac{3}{2}}}
    Hence
    Q43+P43Q^{\dfrac{-4}{3}}+P^{\dfrac{-4}{3}}
    =a43[cos2θ+sin2θ]=a^{-\dfrac{4}{3}}[\cos ^{2}\theta+\sin ^{2}\theta]
    =a43=a^{-\dfrac{4}{3}}.

  • Question 10
    1 / -0
    The equation of the tangent to the curve y=(2x1)e2(1x)\displaystyle y= \left ( 2x-1 \right )e^{2\left ( 1-x \right )} at the point of its maximum is
    Solution
    Given, y=(2x1)e2(1x)y=(2x-1)e ^{2(1-x)}
    y(x)=2(2x1)e2(1x) +2e2(1x) =2e2(1x) (2x+2)\therefore y'\left( x \right) =-2\left( 2x-1 \right) { e }^{ 2\left( 1-x \right)  }+2{ e }^{ 2\left( 1-x \right)  }\\ =2{ e }^{ 2\left( 1-x \right)  }\left( -2x+2 \right)
    Thus y(x)=0x=1y'\left( x \right) =0\Rightarrow x=1 
    Since y(1)<0y''\left( 1 \right) <0 so (1,1)\left( 1,1 \right) is the point of maximum and the equation of tangent is 
    y1=0(x1)y-1=0\left( x-1 \right) ,i.e. y=1y=1

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