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Tangents and its Equations Test 53

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Tangents and its Equations Test 53
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  • Question 1
    1 / -0
    For the curve x=t21x=t^2-1, y=t2ty=t^2-t, the tangent is perpendicular to xx-axis then
    Solution
    Given curve is x=t21,y=t2tx=t^2-1, y=t^2-t

    Derivating w.r.to tt we get

    dxdt=2t\dfrac{dx}{dt}=2t -------(1)

    and dydt=2t1\dfrac{dy}{dt}=2t-1 -------(2)

    dividing (2) by (1)  we get

    dydx=2t12t\dfrac{dy}{dx}=\dfrac{2t-1}{2t}

    Therefore, the slope of the tangent is 2t12t\dfrac{2t-1}{2t}

    Given that the tangent is perpendicular to x-axis. Therefore, tangent is parallel to y-axis. 

    We know that slope of y-axis is infinity and the slopes of the two parallel lines are equal.

    Therefore, slope of the tangent is infinity.

    Hence, 2t12t=10\dfrac{2t-1}{2t}=\dfrac{1}{0}

        2t=0    t=0\implies 2t=0 \implies t=0
  • Question 2
    1 / -0

    Directions For Questions

    Tangent at any point p1p_{1} (other than (0,0)(0, 0)) on the curve y=x3y = x^{3} meets the curve again at p2p_{2}.
    Tangent at p2p_{2} meets the curve again at p3p_{3} and so on.

    ...view full instructions

    Abscissa of p1,p2,p3....pnp_{1}, p_{2}, p_{3} .... p_{n} are in
    Solution

  • Question 3
    1 / -0
    If the line y=4x5y=4x-5 touches to the curve y2=ax3+b{ y }^{ 2 }=a{ x }^{ 3 }+b at the point (2,3)(2,3) then 7a+2b=7a+2b=
    Solution
    The slope of given line y=4x5y=4x-5 is  m=4m=4.

    Now, since the given line touches the curve at (2,3)(2,3), the slope of curve dydx\dfrac{dy}{dx} at the given point is equal to the slope mm of  the line.

    Hence, for slope of curve, differentiating both sides w.r.t x:-

    2ydydx=3ax22y\dfrac{dy}{dx}=3ax^2

    Putting the value x=2x=2 and y=3y=3

    6dydx=3a×4    6×4=12a6\dfrac{dy}{dx}=3a\times 4 \implies 6\times4=12a

         a=2\implies a=2

    Now, the given point lies on the curve, so it must satisfy the equation of the curve:-

        (3)2=2×(2)3+b\implies (3)^2=2\times (2)^3+b

        b=7\implies b=-7

    7a+2b=(7×22×7)=07a+2b=(7\times 2-2\times 7)=0

    Hence, answer is option-(A).
  • Question 4
    1 / -0
    If the tangent at (x1,y1)({x_1},{y_1}) to the curve x3+y3=a3{x^3} + {y^3} = {a^3} meets the curve again at (x2,y2)({x_2},{y_2}), then
    Solution
    Given curve equation is x3+y3=a3x^3+y^3=a^3

    The curve passes through the points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2)

    Therefore, x12+y12=a3x_1^2+y_1^2=a^3 -------(1)

    and x23+y23=a3x_2^3+y_2^3=a^3 -------(2)

    subtracting (1) from (2) we get

    y23y13=(x23x13)y_2^3-y_1^3=-(x_2^3-x_1^3) -------(3)

    Therefore, 3x2+3y2(dydx)=03x^2+3y^2(\dfrac{dy}{dx})=0

        dydx=x2y2\implies \dfrac{dy}{dx}=-\dfrac{x^2}{y^2}

    Therefore, the slope of the tangent at (x1,y1)(x_1,y_1) is x12y12-\dfrac{x_1^2}{y_1^2}

    The tangent passes through the points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2)

    Therefore, the slope of the tangent is y2y1x2x1\dfrac{y_2-y_1}{x_2-x_1}

    On comparing two slopes we get,

    x12y12=y2y1x2x1-\dfrac{x_1^2}{y_1^2}=\dfrac{y_2-y_1}{x_2-x_1}

    we know that a3b3=(ab)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)

        x12y12=y23y13x23x13×x12+x1x2+x22y12+y1y2+y22\implies -\dfrac{x_1^2}{y_1^2}=\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times \dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}

        x12y12=x12+x1x2+x22y12+y1y2+y22\implies -\dfrac{x_1^2}{y_1^2}=-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}

        x12y12+x1x2y12+x22y12=x12y12+x12y1y2+x12y12\implies x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_1^2

        x12y22x22y12=x1x2y12x12y1y2\implies x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2

        (x1y2x2y1)(x1y2+x2y1)=x1y1(x2y1x1y2)\implies (x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)

        x1y2+x2y1=x1y1\implies x-1y_2+x_2y_1=-x_1y_1

        x2x1+y2y1=1\implies \dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1
  • Question 5
    1 / -0
    A point on the curve y=2x3+13x2+5x+9y = 2{x^3} + 13{x^2} + 5x + 9, the tangent at which passes through the origin is 
    Solution
    Let P(a,b)P(a,\,b) be a point on the given curve y=2x3+13x2+5x+9y=2x^3+13x^2+5x+9
    b=2a3+13a2+5a+9b=2a^3+13a^2+5a+9             --------- ( 1 )
    Now, taking derivative of yy.
    \therefore  y=6x2+26x+5yatP=6a2+26a+5y'=6x^2+26x+5\Rightarrow y'\,at\,P=6a^2+26a+5.
    Equation of tangent at PP is
    yb=(6a2+26a+5)(xa)y-b=(6a^2+26a+5)(x-a)
    This tangent passes through origin means
    b=a(6a2+26a+5)b=a(6a^2+26a+5)
    2a3+13a2+5a+9=6a3+26a2+5a2a^3+13a^2+5a+9=6a^3+26a^2+5a
    4a3+13a29=04a^3+13a^2-9=0
    (a+1)(a+3)(4a3)=0(a+1)(a+3)(4a-3)=0
    Taking, a+1=0a+1=0
    We get, a=1a=-1
    Substituting a=1a=-1 in equation (1) we get,
    b=2(1)3+13(1)2+5(1)+9b=2(-1)^3+13(-1)^2+5(-1)+9
    \therefore  b=15b=15
    \therefore  P(a,b)=(1,15)P(a,\,b)=(-1,\,15)
    \therefore   A point on the curve y=2x3+13x2+5x+9y=2x^3+13x^2+5x+9, the tangent at which passes through the origin is (1,15)(-1,\,15).

  • Question 6
    1 / -0
    The curve that passes through the point (2,3)(2,3) and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by
    Solution
    Let BB be the pt. of contact y=f(x)y=f(x) be the curve
    then AB=BCAB=BC (given condition)
    C(2x,0)C\equiv (2x,0) A(0,2y)A\equiv (0,2y)
    Slope mm of line ACAC is 
    m=dydx=0.2y2x0m=\dfrac{dy}{dx}=\dfrac{0.2y}{2x-0}
    dydx=yx\dfrac{dy}{dx}=\dfrac{-y}{x}
    dyy=dxx\dfrac{dy}{y}=\dfrac{-dx}{x}
    lny=lnx+c\ln y=-\ln x+c
    yx=cyx=c
    as (2,3)(2, 3) satisfies the curve
    2×3=c\therefore 2\times 3=c
    y=6x\therefore y=\dfrac{6}{x} is the curve
    \therefore Option CC is correct

  • Question 7
    1 / -0
    The tangent to the curve x=a(θ sinθ);y=a(1+cosθ)x = a(\theta  - \sin \theta );y = a(1 + \cos \theta ) at the points θ =(2k+1)π,kZ\theta  = (2k + 1)\pi ,k \in Z  are parallel to 
    Solution
    x=a(θsinθ),y=a(1+cosθ)dxdθ =a(1cosθ),dydθ =a(sinθ)=asinθdydx=(dydθ  )(dθ dx )=asinθ a(1cosθ)=sinθ 1cosθ \Rightarrow x=a(\theta -\sin\theta ),\quad y=a(1+\cos\theta )\\ \Rightarrow \dfrac { dx }{ d\theta  } =a(1-\cos\theta ),\quad \dfrac { dy }{ d\theta  } =a(-\sin\theta )=-a\sin\theta \\ \therefore \dfrac { dy }{ dx } =\left( \dfrac { dy }{ d\theta  }  \right) \left( \dfrac { d\theta  }{ dx }  \right) =\dfrac { -a\sin\theta  }{ a(1-\cos\theta ) } =\dfrac { -\sin\theta  }{ 1-\cos\theta  }
    \therefore Slope of tangent =sinθ 1cosθ =\dfrac { -\sin\theta  }{ 1-\cos\theta  }
    at θ=(2k+1)π\theta =(2k+1)\pi
    Slope of tangent =sin(2k+1)π 1cos(2k+1)π =0=\dfrac { -\sin(2k+1)\pi  }{ 1-\cos(2k+1)\pi  } =0
    Since the slope of tangent =0=0, it will be parallel to xx- axis
    Therefore y=0y=0 is a possibility.
  • Question 8
    1 / -0
    A chord of parabola y2=4ax{y}^{2}=4ax subtends a right angle at the vertex. The tangents at the extremities of chord intersect on the line:
    Solution

  • Question 9
    1 / -0
    The slope of the tangent to the curve at a point (x,y)(x,y) on it is proportional to (x2).(x-2). If the slope of the tangent to the curve at (10,9)(10,-9)  on it is 3-3. The equation of the curves is .
    Solution

  • Question 10
    1 / -0
    If p1{p_1} and p2{p_2} be the lengths of perpendiculars from the origin on the tangent and normal to the curve x23+y23=a23{x^\frac{2}{3}} + {y^\frac{2}{3}} = {a^\frac{2}{3}}  respectively, then 4p12+p22=4{p_1}^2 + {p_2}^2 =
    Solution

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