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Tangents and its Equations Test 33

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Tangents and its Equations Test 33
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  • Question 1
    1 / -0
    Line y=xy=x and curve y=x2+bx+cy=x^2+bx+c touches at (1,1)(1, 1) then __________.
    Solution
    Given line l:y=xl : y =x
    l:xy=0l : x-y=0
    \therefore slope of line l=1l=1
    Now y=x2+bx+cy=x^2+bx+c
    dydx=2x+b\therefore \dfrac{dy}{dx}=2x+b
    (dydx)P(1,1,)=2(1)+b\therefore \left(\dfrac{dy}{dx}\right)_{P(1, 1,)}=2(1)+b
    =2+b=2+b
    But slope of tangent of the curve =1=1.
    2+b=1\therefore 2+b=1
    b=1\therefore b=-1
    Now (1,1)y=x2+bx+c(1, 1)\in y=x^2+bx+c
    1=1+b(1)+c\therefore 1=1+b(1)+c
    b+c=0\therefore b+c=0
    c=b\therefore c=-b
    c=(1)c=-(-1)
    c=1\therefore c=1
    \therefore We have b=1b=-1 and c=1c=1.
  • Question 2
    1 / -0
    Equation of the normal of curve y=x32x+4y=x^3-2x+4 at point (1,3)(1, 3) is __________.
    Solution
    y=x32x+4y=x^3-2x+4
    dydx=3x22\therefore \dfrac{dy}{dx}=3x^2-2
    (dydx)(1.3)=3(1)2=32=1\therefore \left(\dfrac{dy}{dx}\right)_{(1.3)}=3(1)-2=3-2=1
    \therefore slope of tangent =1=1
    \therefore slope of normal =1=-1
    \therefore We get equation of Normal using
    uy1=m(xx1)u-y_1=m(x-x_1)
    y3=1(x1)\therefore y-3=-1(x-1)
    y3=x+1\therefore y-3=-x+1
    x+y31=0\therefore x+y-3-1=0
    \therefore Required equation of Normal is x+y4=0x+y-4=0
  • Question 3
    1 / -0
    The equation to the tangent to (xa)n+(yb)n=2{\left( {{\dfrac xa}} \right)^n} + {\left( {{\dfrac yb}} \right)^n} = 2 at (a,b)\left( {a,b} \right)


    Solution
    (xa)n+(yb)nn(xa)n1+n(yb)n1dydx=0{ (\dfrac { x }{ a } ) }^{ n }+{ (\dfrac { y }{ b } ) }^{ n }\\ n{ (\dfrac { x }{ a } ) }^{ n-1 }+n{ (\dfrac { y }{ b } ) }^{ n-1 }\dfrac { dy }{ dx } =0
    At point (a,b)(a,b)
    x=a , y=bx=a\ ,\ y=b
    n+ndydx=0dydx=1=my=mx+cy=x+cn+n\dfrac { dy }{ dx } =0\\ \dfrac { dy }{ dx } =-1=m\\ y=mx+c\\ y=-x+c
    at(a,b)(a,b)
    b=a+cb=-a+c
    c=b+ac=b+a
    y+x=b+ay+x=b+a

  • Question 4
    1 / -0
    Find the equations of the tangent and the normal to the curve x2+y2=5x^2 +y^2=5 where the tangent is parallel to the lines xy=0x-y =0 respectively.
    Solution
    Let the tangent equation be xy=cx-y=c

    Substituting above in circle equation ;

    (c+y)2+y2=5\Rightarrow (c+y) ^{2}+y^{2}=5

    2y2+2cy+c25=0\Rightarrow 2y^{2}+2cy+c^{2}-5=0

    As tangent intersects at only single point, above equation should have equal roots 

    \Rightarrow Discriminant =0=0

    4c2=4×2×(c25)\Rightarrow 4c^{2}=4×2×(c^{2}-5)

    c=±10\Rightarrow c=\pm \sqrt {10}

    \therefore Equation of tangent is : xy=±10x-y=\pm \sqrt {10}

    As normal is perpendicular to tangent, let the equation of normal be x+y=kx+y=k

    And normal passes through the center of circle (0,0)(0,0)

    0+0=k\Rightarrow 0+0=k

    k=0\Rightarrow k=0

    \therefore Equation of normal is x+y=0x+y=0
  • Question 5
    1 / -0
    The point on the curve y2=xy^{2}=x where tangent makes 45o45^{o} angle with xx-axis ?
    Solution
    dydx= \dfrac{dy}{dx} = Slope of the tangent at that point of the curve.

    Given curve is x=y2 x = y^2
    Keep y=a,then x=y2 y = a, \, then \,  x = y^2  
    point (a2,a) (a^2, a)

    dydx(y2x)=0dydx=12y=12a=tan(450) \displaystyle \dfrac{dy}{dx} (y^2 - x) = 0 \dfrac{dy}{dx} = \dfrac{1}{2y} = \dfrac{1}{2a} = \tan(45^0)

    a=12a2=14 \Rightarrow a= \dfrac{1}{2} \quad a^2= \dfrac{1}{4}

    point (14,12) \left(\dfrac{1}{4}, \dfrac{1}{2}\right)
  • Question 6
    1 / -0
    Equation of the normal to the curve y=x+2y = -\sqrt{x} + 2 at the point of its interaction with the curve y=xy = x is
    Solution
    Equation of normal to the curve
    y=x+2y=-\sqrt{x}+2 at the point of intersection with y=xy=x.
    Point of intersection of
    y=x+2y=-\sqrt{x}+2, y=xy=x
    x=x+1x=-\sqrt{x}+1
    x2=xx-2=-\sqrt{x}
    x24x+4=xx^2-4x+4=x
    x25x+4=0x^2-5x+4=0
    (x4)(x1)=0(x-4)(x-1)=0
    x=4,1x=4, 1
    If x=4x=4, (4,4)(4, 4) does not
    Satisfy y=x+2y=-\sqrt{x}+2
    So, intersection point is (1,1)(1, 1)
    Normal at (1,1)(1, 1) to y=x+2y=-\sqrt{x}+2
    dydx=12x\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{x}}
    dydx(1,1)=12\dfrac{dy}{dx}|_{(1, 1)}=\dfrac{-1}{2}
    Slope of normal =(1)(12)=2=\dfrac{(-1)}{\left(\dfrac{-1}{2}\right)}=2 and passes through (1,1)(1, 1)
    Equation of normal: 2xy=12x-y=1.
  • Question 7
    1 / -0
    The perpendicular distance from origin to the normal at any point to the curve x=a(cosθ +θsinθ)x = a\left( {\cos \theta  + \theta \sin \theta } \right)y=a(sinθ θcosθ)y = a\left( {\sin \theta  - \theta \cos \theta } \right)
    Solution
    dydx=dydθ dθdx =tanθ= slope of tangent\dfrac{dy}{dx} = \dfrac{dy}{d\theta}  \dfrac{d\theta}{dx}  = tan \theta = \space slope\space of\space tangent

    Slope of the normal =cot θ=-\cot \space \theta =tan(90+θ)=\tan(90+\theta)

    [ya(sinθθcosθ)]=cosθsinθ(xa(cosθ+asinθ))[y-a(\sin\theta - \theta \cos \theta)] = \dfrac{-\cos\theta}{\sin\theta} (x-a(\cos\theta+a \sin\theta))

    xcosθ+ysinθ=a(1)x \cos\theta +y \sin\theta = a(1)

    Now distance from (0,0)(0,0)

    d=(0+0a)1=ad=\dfrac{(0+0-a)}{1} =a

  • Question 8
    1 / -0
    The slope of the tangent and normal to y=x23x+5y={ x }^{ 2 }-3x+5 at (2,3)(2,3) are ______ and _____ respectively.
    Solution
    the slope of the tangent at any point is the derivative of the equation at that point

    therefore

    ddx(x23x+5)=2x3\frac { d }{ dx } ({ x }^{ 2 }-3x+5)=2x-3

    at(2,3)(2,3) dydx\dfrac{dy}{dx} is 2(2)3=12(2)-3=1

    therefore the slope of the tangent is 11

    normal is perpendicular to the tangent

    therefore the slope of the normal*slope of the tangent=1-1

    therefore the slope of the normal is 1-1

  • Question 9
    1 / -0
    Find the angle between tangent of the curve y=(x+1)(x3)y = (x + 1) (x - 3) at the point where it cuts the axis of xx.
    Solution
    The curve y=(x+1)(x3)=x22x3y=(x+1)(x-3)=x^2-2x-3.........(1) cuts the axis of xx at (1,0)(-1,0) and (3.0)(3.0). These points are obtained by putting y=0y=0 in the equation (1).
    Now the slope of the tangent to the curve (1) at (1,0)(-1,0) be (m1)=dydx(1,0)=[2x2](1,0)=4(m_1)=\left.\dfrac{dy}{dx}\right|_{(-1,0)}=[2x-2]_{(-1,0)}=-4.
    Again the slope of the tangent to the curve (1) at (1,0)(-1,0) be (m2)=dydx(3,0)=[2x2](3,0)=4(m_2)=\left.\dfrac{dy}{dx}\right|_{(3,0)}=[2x-2]_{(3,0)}=4.
    Now the angle between the tangents to (1) at those points be 
    tan1(m1m21+m1.m2)\tan^{-1}\left(\dfrac{m_1-m_2}{1+m_1.m_2}\right)

    =tan1(815)=\tan^{-1}\left(\dfrac{8}{15}\right).
    So the required angle be tan1(815)\tan^{-1}\left(\dfrac{8}{15}\right).
  • Question 10
    1 / -0
    If the tangent at the P of the curve y2=x3{y^2} = {x^3} intersect the curve again at Q and the straight lines OP, OQ make angles α,β\alpha ,\beta with the x-axis, where O is the origin. then, tanα/tanβ{{\tan \alpha } / {\tan \beta }} has the value equal to 
    Solution
    Let the parametric coordinates at PP and QQ be :
    (t12,t13),(t22,t23)(t_1^2, t_1^3), (t_2^2, t_2^3)
    Then slope of the tangent at the point Pdydx=3x22y=32t1P\Rightarrow \dfrac{dy}{dx}=\dfrac{3x^2}{2y}=\dfrac{3}{2t_1}
    Also, line joining PP and QQ would have a slope given by: t23t13t22t12\dfrac{t_2^3-t_1^3}{t_2^2-t_1^2}.
    Equating both slopes, 2t22t12=t2t1+12\dfrac{t_2^2}{t_1^2}=\dfrac{t_2}{t_1}+1------(1)
    Also. tanα=t1,tanβ=t2\tan \alpha=t_1, \tan \beta=t_2-------(2)
    \therefore From (1),
    2tan2βtan2α=tanβtanα+12\dfrac{\tan^2\beta}{\tan^2\alpha}=\dfrac {\tan \beta}{\tan \alpha}+1.
    Let t2t1=x\dfrac{t_2}{t_1}=x
    2x2=x+1\therefore 2x^2=x+1
    or, 2x2x1=02x^2-x-1=0
    or, 2x22x+x1=02x^2-2x+x-1=0
    or, 2x(x1)+1(x1)=0(x1)(2x+1)=02x(x-1)+1(x-1)=0\Rightarrow (x-1)(2x+1)=0
    x=1,1/2\therefore \boxed{x=1, 1/2}
    i.e t2t1=1t2=t1\dfrac{t_2}{t_1}=1\Rightarrow \boxed{t_2=t_1} or, t2t1=12t1=2t2\dfrac{t_2}{t_1}=\dfrac{-1}{2}\Rightarrow \boxed{t_1=-2t_2}
    \therefore From (2) if t2=t1t_2=t_1;    and if t1=2t2t_1=2t_2
    tanα=tanβ\tan \alpha=\tan \beta                        tanα=2tanβ\tan \alpha=-2\tan \beta
    tanαtanβ=1\therefore \boxed{\dfrac{\tan \alpha}{\tan \beta}=1}                    tanαtanβ=2\therefore \boxed{\dfrac{\tan \alpha}{\tan \beta}=-2}

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