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Tangents and its Equations Test 15

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Tangents and its Equations Test 15
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  • Question 1
    1 / -0
    If the tangent at each point of the curve $$\displaystyle y=\frac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$$ makes an acute angle with the positive direction of x-axis, then 
    Solution
    We have, $$\displaystyle  y=\dfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$$ 
    $$\displaystyle \Rightarrow \dfrac { dy }{ dx } =2{ x }^{ 2 }-4ax+2.$$
    Since, the tangent makes an acute angle with the positive direction of x-axis, therefore, 
    $$\displaystyle \dfrac { dy }{ dx } \ge 0\Rightarrow 2{ x }^{ 2 }-4ax+2\ge 0$$ for all $$x$$
    $$\Rightarrow 16{ a }^{ 2 }-16\le 0$$
    $$(\because $$ Disc. $$={ \left( 4a \right)  }^{ 2 }-4\left( 2 \right) \left( 2 \right) \le 0)$$
    $$\Rightarrow { a }^{ 2 }-1\le 0$$ i.e., $$\left( a-1 \right) \left( a+1 \right) \le 0$$
    $$\Rightarrow -1\le a\le 1.$$ 
  • Question 2
    1 / -0
    If a variable tangent to the curve $$\displaystyle x^2y = c^3$$ makes intercepts a, b on x and y axis respectively, then the value of $$\displaystyle a^2b$$ is
    Solution
    Let a variable point$$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$  on the curve $${ x }^{ 2 }y={ c }^{ 3 }$$
    Then slope $$\cfrac { dy }{ dx } =-\cfrac { 2{ y }_{ 1 } }{ { x }_{ 1 } } $$
    So, equation of tangent is 
    $$y-{ y }_{ 1 }=-\cfrac { 2{ y }_{ 1 } }{ { x }_{ 1 } } \left( x-{ x }_{ 1 } \right) $$
    $$y{ x }_{ 1 }-2x{ y }_{ 1 }=3{ x }_{ 1 }{ y }_{ 1 }$$
    Using $${ { x }_{ 1 } }^{ 2 }{ y }_{ 1 }={ c }^{ 3 }$$
    Equation of tangent becomes $$y{ x }_{ 1 }^{ 2 }+\cfrac { 2{ c }^{ 3 }x }{ { x }_{ 1 } } =3{ c }^{ 3 }$$
    Therefore, X intercept is $$a=\cfrac { 3{ x }_{ 1 } }{ 2 } $$ 
    And Y intercept is $$b=\cfrac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } } $$

    Now $${ a }^{ 2 }b={ \left( \cfrac { 3{ x }_{ 1 } }{ 2 }  \right)  }^{ 2 }\left( \cfrac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } }  \right) =\cfrac { 27{ c }^{ 3 } }{ 4 } $$
  • Question 3
    1 / -0
    If $$y = 4x - 5$$ is a tangent to the curve $$\displaystyle y^2 = px^3 + q$$ at $$(2, 3)$$, then
    Solution
    Given equation of curve
    $$\displaystyle y^2 = px^3 + q\\$$
    $$\displaystyle \dfrac{dy}{dx}=\dfrac{3px^{2}}{y}\\$$
    Slope of tangent at (2,3) $$= \displaystyle (\dfrac{dy}{dx})_{(2,3)}=2p$$

    Given equation of tangent is 
    $$y=4x-5$$
    Slope of tangent =4

    $$\Rightarrow 2p=4$$
    $$\Rightarrow p=2$$

    Since, (2,3) lies on 
    $$ y^{2} = px^{3} + q$$
    $$9=16+q$$
    $$\Rightarrow q=-7$$
  • Question 4
    1 / -0
    The slope of the tangent to the curve $$y=x^{2}-x$$ at the point where the line $$y=2$$ cuts the curve in the first quadrant is
    Solution
    For $$y=2, { x }^{ 2 }-x-2=0$$ gives $$x=-1$$ and $$x=2$$ as point is in first quadrant.
    Therefore point is $$\left( 2,2 \right) $$
    Now $$\cfrac { dy }{ dx } =2x-1=4-1=3$$
  • Question 5
    1 / -0
    If the tangent to the curve $$xy + ax + by = 0$$ at (1, 1) makes an angle $$\displaystyle \tan ^{-1}(2)$$ with x-axis, then $$\displaystyle a + 2b$$ is equal to
    Solution
    $$xy'+y+a+by'=0$$
    $$(x+b)y'+a+y=0$$
    $$y'=\dfrac{-(a+y)}{x+b}$$
    Now 
    $$y'_{1,1}=\dfrac{-(a+1)}{b+1}$$
    $$=2$$
    Or 
    $$2b+2=-a-1$$
    $$2b+a=-3$$
  • Question 6
    1 / -0
    If at each point of the curve $$y=x^{3}-ax^{2}+x+1$$ the tangent is inclined at an acute angle with the positive direction of the x-axis then
  • Question 7
    1 / -0
    The slope of the tangent to the curve $$y=\sqrt{4-x^{2}}$$ at the point where the ordinate and the abscissa are equal, is
    Solution
    Let abscissa = ordinate = $$a$$ , then point is $$\left( a,a \right) $$
    Now for $$y=\sqrt { 4-{ x }^{ 2 } } \Rightarrow { y }^{ 2 }=4-{ x }^{ 2 }$$
    Hence slope is $$\cfrac { dy }{ dx } =\cfrac { -2x }{ 2y } =\cfrac { -2a }{ 2a } =-1$$
  • Question 8
    1 / -0
    If the tangent to the curve $${ 2y }^{ 3 }={ ax }^{ 2 }+{ x }^{ 3 }$$ at the point $$(a,a)$$ cuts off intercepts $$\alpha $$ and $$\beta $$ on the coordinate axes such that $${ \alpha  }^{ 2 }+{ \beta  }^{ 2 }=61,$$ then $$a=$$
    Solution
    The slope of the tangent at any point $$(a,a)$$ on the curve is $$\displaystyle { m }_{ T }={ \frac { dy }{ dx }  }_{ \left( a,a \right)  }$$

    $$\displaystyle \therefore { 6y }^{ 2 }\frac { dy }{ dx } =2ax+{ 3x }^{ 2 }$$

    $$\displaystyle \Rightarrow \frac { dy }{ dx } =\frac { 2ax+{ 3x }^{ 2 } }{ { 6y }^{ 2 } } $$

    $$ \displaystyle \Rightarrow { m }_{ T }={ \frac { dy }{ dx }  }_{ \left( a,a \right)  }=\frac { 5 }{ 6 } $$ 

    The equation of the tangent at $$(a,a)$$ is $$\displaystyle y-a=\frac { 5 }{ 6 } \left( x-a \right) \Rightarrow 5x-6y+a=0$$
    This cuts off intercepts of lengths $$\displaystyle -\frac { a }{ 5 } $$ and 

    $$\displaystyle \frac { a }{ 6 } $$ with $$x$$ and $$y$$-axis respectively. 
    So, $$\displaystyle \alpha =-\frac { a }{ 5 } $$ and $$\displaystyle \beta =\frac { a }{ 6 } .$$
    Since, $${ \alpha  }^{ 2 }+{ \beta  }^{ 2 }=61$$ {given}

    $$\displaystyle \therefore \frac { a^{ 2 } }{ 25 } +\frac { { a }^{ 2 } }{ 36 } =61\Rightarrow { a }^{ 2 }=25\times 36\therefore a=\pm 30.$$
  • Question 9
    1 / -0
    If $$m$$ be the slope of a tangent to the curve $${ e }^{ 2y }=1+4{ x }^{ 2 }$$, then 
    Solution
    We have, $$\displaystyle { e }^{ 2y }=1+4{ x }^{ 2 }\quad \Rightarrow { e }^{ 2y }.2\dfrac { dy }{ dx } =8x$$
    $$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac { 4x }{ { e }^{ 2y } } =\dfrac { 4x }{ 1+4{ x }^{ 2 } } .$$
    $$\therefore$$ Slope of tangent $$\displaystyle =m=\dfrac { 4x }{ 1+4{ x }^{ 2 } } $$
    $$\displaystyle \Rightarrow \left| m \right| =\dfrac { 4\left| x \right|  }{ 1+4{ \left| x \right|  }^{ 2 } } \le 1$$
    $$\displaystyle \left[ \because { \left( 1-2\left| x \right|  \right)  }^{ 2 }\ge 0\quad \quad \Rightarrow 1+4{ \left| x \right|  }^{ 2 }-4\left| x \right| \ge 0 \Rightarrow \dfrac { 4\left| x \right|  }{ 1+4{ \left| x \right|  }^{ 2 } } \le 1 \right] $$
  • Question 10
    1 / -0
    If $$m$$ be the slope of tangent to the curve $$e^{y}=1+x^{2}$$ then 
    Solution
    $$e^{y}=1+x^{2}$$
    Or 
    $$y=ln(1+x^{2})$$
    $$\dfrac{dy}{dx}$$
    $$=m$$
    $$=\dfrac{2x}{1+x^{2}}$$
    Now 
    $$(1-x)^{2}\geq0$$
    Or 
    $$1+x^{2}-2x\geq0$$
    Or 
    $$1+x^{2}\geq2x$$
    Or 
    $$\dfrac{2x}{1+x^{2}}\leq 1$$
    Hence
    $$|m|\leq 1$$
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