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Tangents and its Equations Test 38

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Tangents and its Equations Test 38
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  • Question 1
    1 / -0
    The normal to the curve, $${x}^{2}+2xy-{3y}^{2}=0,\ at\left (1,1\right)$$:
    Solution

  • Question 2
    1 / -0
    If the tangent at P of the curve $$y^2=x^3$$ intersects the curve again at Q and the straight lines OP, OQ ma angles $$\alpha, \beta$$ with the x-axis where 'O' is the origin then $$\tan\alpha/\tan\beta$$ has the value equal to?
    Solution
    Let $$P$$ be $$(x_{1}, y_{1})$$ and $$Q$$ be $$(x_{2}, y_{2})$$
    equation of tangent $$(x_{2}, y_{2})$$, so $$\dfrac{y_{0}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$$
    need to find out $$\dfrac{\tan \alpha}{ \tan \beta}  = \dfrac{\dfrac{y_{1}}{x_{1}}}{y_{2}/x_{2}} = \dfrac{y_{1}x_{2}}{x_{1} y_{2}}$$
    $$\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$$
    $$2y_{1}y_{2}-2y^{2}= 3x^{2}_{1}x_{2}-3x_{1}^{3}$$
    $$2y_{1}y_{2}-2y_{1}^{2}= 3x_{1}^{2}x_{2}-3y^{2}$$
    $$2y_{1}y_{2}+y_{1}^{2}= 3x_{1}^{2} x_{2}$$
    $$y_{1}^{3} (2y_{2}+y_{1})^{3}= 27 x_{1}^{6} x^{3}_{2}$$
    $$y_{1}^{3} (2y_{2}+ y_{1})^{3}= 27 y_{1}^{4} y_{2}^{3}$$
    $$(2y_{2}+y_{1})^{3} - 27 y_{1} y_{2}$$
    $$8 y_{2}^{3} - 15 y_{2}^{2} y_{1}+ 6y_{2} y^{2}_{1}+ y_{1}^{3}=0$$
    $$(y_{2}-y_{1})^{2} (8y_{2}+ y_{1})=0$$
    $$8y_{2}=-y_{1} \Rightarrow 64 y_{2}^{2} = y_{1}^{2}$$
    $$64 x_{2}^{3} = x_{1}^{3} \Rightarrow 4 x_{2} = x_{1}$$
    $$\dfrac{\tan \alpha}{\tan \beta}= \dfrac{y_{1}x_{2}}{x_{1}y_{2}}$$
    $$= \dfrac{-8y_{2} x_{2}}{4x_{2} y_{2}} = -2$$
  • Question 3
    1 / -0
    If $$x={t}^{2}$$ and $$y=2t$$, then equation of the normal at $$t=1$$ is
    Solution
    $$x=t^2$$,$$y=2t$$
    $$y^2=4t^2=4x$$               $$t=1$$
    $$y^2=4x$$                         $$x=1,y=2$$
    $$2y \dfrac{dy}{dx}=4 \rightarrow \dfrac{dy}{dx}=\dfrac{2y}{y}$$
    $$y=2$$
    $$\dfrac{dy}{dx}=1$$
    $$-\dfrac{dx}{dy}=-1$$
    equation of normal
    $$y-2=-1(x-1)$$
    $$y-2+x-1=0$$
    $$x+y-3=0$$

  • Question 4
    1 / -0
    The area of the triangle formed by the coordinate axes and a tangent to the curve $$xy={a}^{2}$$ at the point $$({x}_{1},{y}_{1})$$ is
    Solution
    $$\dfrac { dy }{ dx } =\dfrac { { -a }^{ 2 } }{ { x }^{ 2 } }$$
    $${ x }_{ 1 }{ y }_{ 1 }={ a }^{ 2 }$$
    $$y-{ y }_{ 1 }=\dfrac { { -a }^{ 2 } }{ { x }_{ 1 }^{ 2 } } \left( x-{ x }_{ 1 } \right)$$
    $$x=0;y-{ y }_{ 1 }=\dfrac { { +a }^{ 2 } }{ { x }_{ 1 }^{ 2 } } \left( 0-{ x }_{ 1 } \right)$$
    $$x=0;y={ y }_{ 1 }+\dfrac { { +a }^{ 2 } }{ { x }_{ 1 } }$$
    $$y=0;{ +y }_{ 1 }=\dfrac { { +a }^{ 2 } }{ { x }_{ 1 }^{ 2 } } \left( x-{ x }_{ 1 } \right)$$
    $$x={ x }_{ 1 }+\dfrac { { x }_{ 1 }^{ 2 }{ y }_{ 1 } }{ { a }^{ 2 } }$$
    Area $$=\dfrac { 1 }{ 2 } \left( { x }_{ 1 }+\dfrac { { x }_{ 1 }^{ 2 }{ y }_{ 1 } }{ { a }^{ 2 } }  \right) \left( { y }_{ 1 }+\dfrac { { a }^{ 2 } }{ { x }_{ 1 } }  \right)$$
    $$=\dfrac { 1 }{ 2 } \left( { x }_{ 1 }{ y }_{ 1 }+{ a }^{ 2 }+\dfrac { { \left( { x }_{ 1 }{ y }_{ 1 } \right)  }^{ 2 } }{ { a }^{ 2 } } +{ x }_{ 1 }{ y }_{ 1 } \right)$$
    $$=\dfrac { 1 }{ 2 } \left( { a }^{ 2 }+{ a }^{ 2 }+\dfrac { { a }^{ 4 } }{ { a }^{ 2 } } +{ a }^{ 2 } \right)$$
    $$=\dfrac { 4{ a }^{ 2 } }{ 2 } =2{ a }^{ 2 }$$

  • Question 5
    1 / -0
    The equation of the tangent to curve $$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=2$$ at the point (a, b) is 
    Solution

    We have,

    $$ \sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}}=2\,\,.......\,\,\left( 1 \right) $$

    $$ \dfrac{\sqrt{x}}{\sqrt{a}}+\dfrac{\sqrt{y}}{\sqrt{b}}=2 $$

    On differentiation and we get,

    $$ \dfrac{1}{2\sqrt{x}\sqrt{a}}+\dfrac{1}{2\sqrt{y}\sqrt{b}}\dfrac{dy}{dx}=0 $$

    $$ \dfrac{1}{2\sqrt{y}\sqrt{b}}\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}\sqrt{a}} $$

    $$ \dfrac{dy}{dx}=-\dfrac{\sqrt{y}\sqrt{b}}{\sqrt{x}\sqrt{a}} $$

    At the point $$\left( a,\,b \right)$$ and we get,

    $$ \dfrac{dy}{dx}=-\dfrac{\sqrt{b}\sqrt{b}}{\sqrt{a}\sqrt{a}} $$

    $$ \dfrac{dy}{dx}=-\dfrac{b}{a} $$

    Equation of tangent is

    $$ y-{{y}_{1}}=\dfrac{dy}{dx}\left( x-{{x}_{1}} \right) $$

    $$ y-b=-\dfrac{b}{a}\left( x-a \right) $$

    $$ ay-ab=-bx+ab $$

    $$ ay+bx=2ab $$

    $$ \dfrac{ay+bx}{ab}=2 $$

    $$ \dfrac{y}{b}+\dfrac{x}{a}=2 $$

    $$ \dfrac{x}{a}+\dfrac{y}{b}=2 $$

    Hence, this is the answer.
  • Question 6
    1 / -0
    The equation of tangent to the ellipse $$4{x^2} + 9{y^2} = 36$$ at $$(3, - 2)\,is\,$$
    Solution
    The equation of tangent at $$(x_1,y_1)$$ to the ellipse $$4{x^2} + 9{y^2} = 36$$ is $$4xx_1+9yy_1=36$$......(1).
    Here $$x_1=3, y_1=-2$$.
    Then the equaiton of the tangent will be $$4x.3+9y.(-2)=36$$ or, $$2x-3y=6$$.
  • Question 7
    1 / -0
    The equation of tangents to the ellipse $${x^2} + 4{y^2} = 25$$ at the point whose ordinate is 2, is 
    Solution
      $$ Given $$
     $$ {{x}^{2}}+4{{y}^{2}}=25 $$
     $$ formula\,used $$
     $$ equation\text{ of tangent from point}\left( x_{1}^{{}},y_{1}^{{}} \right)to\,\,ellipse\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\,is $$
     $$ \frac{xx_{1}^{{}}}{{{a}^{2}}}+\frac{yy_{1}^{{}}}{{{b}^{2}}}=1 $$
     $$ {{x}^{2}}+4{{y}^{2}}=25 $$
     $$ \Rightarrow \frac{{{x}^{2}}}{25}+\frac{4{{y}^{2}}}{25}=1 $$
     $$ let\,the\,po\operatorname{int}\,be\,\left( x,2 \right) $$
     $$ hence $$
     $$ \frac{{{x}^{2}}}{25}+\frac{4{{\left( 1 \right)}^{2}}}{25}=1 $$
     $$ \Rightarrow \frac{{{x}^{2}}}{25}=1-\frac{16}{25} $$
     $$ \Rightarrow {{x}^{2}}=9 $$
     $$ \Rightarrow x=\pm 3 $$
     $$ Hence\,the\,po\operatorname{int}\,is\,\left( -3,2 \right),\left( 3,2 \right) $$
     $$ equation\,of\,\tan gent\,at\left( 3,2 \right) $$
     $$ \frac{3x}{25}+\frac{4\left( 2y \right)}{25}=1 $$
     $$ \Rightarrow 3x+8y=25 $$
  • Question 8
    1 / -0
    The equation of the normal to the curve $${x}^{4}=4y$$ through the point $$(2,4)$$ is
    Solution
    $$\dfrac {m}{4}=y$$
    $$\dfrac {dy}{dx}=\dfrac {4x^3}{4}=x^3$$
    $$\dfrac {dy}{dx}|_{(2,4)} =2^3=8$$
    lquaton of normal $$\rightarrow y-y_1=\dfrac {-1}{dy/dx}(x-x_1)$$
    $$y-4=\dfrac {-1}{8}(x-2)$$
    $$8y-32=-x+2$$
    $$x+8y=34$$

  • Question 9
    1 / -0
    Number of tangents drawn from the point $$\left (-1/2,0\right)$$ to the curve $$y={e}^{x}$$. (Here { } denotes fractional part function ). 
    Solution
    Let $$A$$ be the point of contact of a tangent of $${e}^{x}$$ passing through $$(-\cfrac{1}{2},0)$$
    $$\Rightarrow$$ $${ \left| \cfrac { dy }{ dx }  \right|  }_{ \lambda  }={ e }^{ \lambda  }$$
    Equation of tangent $$\quad y-{ e }^{ \lambda  }={ e }^{ \lambda  }(x-\lambda )\quad $$
    $$x=-\cfrac { 1 }{ 2 } ,y=0$$
    $$-{ e }^{ \lambda  }={ e }^{ \lambda  }\left( -\cfrac { 1 }{ 2 } -\lambda  \right) \Rightarrow { e }^{ \lambda  }\lambda =\cfrac { { e }^{ \lambda  } }{ 2 } $$
    either
    $$\quad { e }^{ \lambda  }=0$$ or $$\lambda =+\cfrac { 1 }{ 2 } $$
    $$\quad y{ e }^{ \lambda  }=0\Rightarrow y=0\quad \rightarrow$$ xaxis
    $$\quad \lambda =\cfrac { 1 }{ 2 } \Rightarrow y-{ e }^{ \cfrac { 1 }{ 2 }  }={ e }^{ \cfrac { 1 }{ 2 }  }\left( x-\cfrac { 1 }{ 2 }  \right) $$ 
    2 solutions or 2 distinct tangents
  • Question 10
    1 / -0
    The equation of tangent to the curve $$y=x^{2}+4x+1$$ at $$\left(-1,-2\right)$$ is
    Solution
    $$\dfrac{dy}{dx}$$ at $$(-1,-2)$$ be:-
    $$\dfrac{d}{dx}(x^{2}+4x+1)=2x+4$$
    $$2(-1)+4=2$$
    $$\therefore $$ Equation of tangent : - 
    $$y -(-2)=2(x-(-1))$$
    $$\therefore y +2=2x+2$$
    $$\therefore y =2x$$
    $$\therefore 2x-y=0$$
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