Tangents and its Equations Test 1

As $$\left| f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right)  \right| \le { \left( { x }_{ 1 }-{ x }_{ 2 } \right)  }^{ 2 },\forall { x }_{ 1 },{ x }_{ 2 }\in R$$
$$\displaystyle\Rightarrow \left| f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right)  \right| \le { \left| { x }_{ 1 }-{ x }_{ 2 } \right|  }^{ 2 }$$   $$\left( { x }^{ 2 }={ \left| x \right|  }^{ 2 } \right) $$
$$\displaystyle \therefore \left| \frac { f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right)  }{ { x }_{ 1 }-{ x }_{ 2 } }  \right| \le { \left| { x }_{ 1 }-{ x }_{ 2 } \right|  }\Rightarrow \lim _{ { x }_{ 1 }\rightarrow { x }_{ 2 } }{ \left| \frac { f\left( { x }_{ 1 } \right) -f\left( { x }_{ 2 } \right)  }{ { x }_{ 1 }-{ x }_{ 2 } }  \right|  } \le \lim _{ { x }_{ 1 }\rightarrow { x }_{ 2 } }{ \left| { x }_{ 1 }-{ x }_{ 2 } \right|  }$$
$$\displaystyle\Rightarrow \left| f'\left( { x }_{ 1 } \right)  \right| \le 0,\forall { x }_{ 1 }\in R$$
$$\therefore \left| f'\left( x \right)  \right| \le0$$, which showsw $$\left| f'\left( x \right)  \right| =0$$   (as modulus is non negative or $$\left| f'\left( x \right)  \right| \ge 0$$)
$$\therefore f'\left( x \right)=0$$  or $$f\left( x \right)$$ is a constant function.
$$\Rightarrow $$ Equation of tangent at $$\left( 1,2 \right) $$ is 
$$\displaystyle \frac { y-2 }{ x-1 } =f'\left( x \right)$$ or $$y-2=0$$ [$$\because$$ as $$f'(x)=0$$]
$$\Rightarrow y-2=0$$ is required equation of tangent.

$$y=3\sin { \theta  } \cos { \theta  } $$

$$x^2y^2-2x=4-4y$$
$$2xy^2+2y\cdot x^2\cdot\displaystyle\frac{dy}{dx}-2=-4\cdot\frac{dy}{dx}$$
$$\displaystyle\frac{dy}{dx}(2y\cdot x^2+4)=2-2x\cdot y^2$$
$$\left.\displaystyle\frac{dy}{dx}\right|_{2, -2}=\displaystyle\frac{2-2\times 2\times 4}{2(-2)\times 4+4}=\frac{+14}{+12}=\frac{7}{6}$$
 eq of tangent :$$(y+2)=\displaystyle\frac{7}{6}(x-2)\Rightarrow 7x-6y=26$$.

Clearly $$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\tan\theta =$$ slope of normal $$=-\cot\theta$$
Equation of normal at $$\theta$$' is $$
\mathrm{y}-\mathrm{a}(\sin\theta -\theta \cos\theta)=-\cot\theta(\mathrm{x}-\mathrm{a}(\cos\theta +\theta \sin\theta)$$
$$=$$ $$y\sin{\theta}-a\sin^{2}\theta +a\theta \cos\theta \sin\theta =-\mathrm{x}\cos\theta +a\cos^{2}\theta +a\theta \sin\theta \cos\theta$$
$$= x\cos{\theta} + y\sin{\theta} =a$$
Clearly this is an equation of straight line which is at a constant distance $$a$$ from origin. 

Given equation of curve is $$x^{2}=y-6$$

Given equation of conic is
$$y - 6 = x^2$$ 
$$\dfrac{dy}{dx}=2x$$
Slope of tangent at $$(2,10)$$ is $$4.$$

Equation of tangent to conic is 
$$y-10=4(x-2)$$
$$\Rightarrow y-10=4x-8$$
$$\Rightarrow y=4x+2$$          .....(1)

Given equation of circle is 
$$x^{ 2 }+y^{ 2 }+8x-2y=k$$
$$\Rightarrow (x+4)^2+(y-1)^2=k+17$$

Given $$(\alpha, \beta)$$ is a point of tangency for fixed $$k$$
Radius = Length of tangent from center $$(-4,1)$$
$$\Rightarrow \sqrt{k+17}=\dfrac{15}{\sqrt{17}}$$
$$\Rightarrow k+17=\dfrac{225}{17}$$

So, equation of circle at $$(\alpha, \beta)$$ is 
$$(\alpha+4)^2+(\beta-1)^2=\dfrac{225}{17}$$

Now, eqn (1) is tangent to the circle for fixed $$k$$ at $$(\alpha,\beta)$$
$$\Rightarrow (\alpha+4)^2+(4\alpha+1)^2=\dfrac{225}{17}$$
$$\Rightarrow 17\alpha^2+16\alpha+17=\dfrac{225}{17}$$
$$\Rightarrow 289\alpha^2+272\alpha+64=0$$
$$\Rightarrow \alpha =\dfrac{-8}{17}$$  (Here $$D=0$$)

So, by (1), $$\beta =\dfrac{2}{17}$$

$$\dfrac{dy}{dx}=|x|=2\Rightarrow  x=\pm 2\Rightarrow  y=\int_{0}^{2}|t|$$ dt $$=2$$ for $$x=2$$
and $$y=\int_{0}^{-2}|t|$$ dt $$=-2$$ for $$x=-2$$
Hence the equations of the tangents are
$$y-2=2(x-2) \Rightarrow  y=2x-2$$
and $$y+2=2(x+2) \Rightarrow  y=2x+2$$
Putting $$y=0$$, we get $$x=1$$ and $$-1$$.