Tangents and its Equations Test 24

$$\displaystyle y=\cos \left ( x+y \right )$$           $$\displaystyle \left ( -2\pi \leq x\leq 2\pi  \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\sin \left ( x+y \right )\left ( 1+\frac{dy}{dx} \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}\left ( 1+\sin \left ( x+y \right ) \right )=-\sin \left ( x+y \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}$$
Given tangent is parallel to the line, $$x + 2y = 0$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{\sin \left ( x+y \right )}{1+\sin \left ( x+y \right )}=-\frac{1}{2}$$
$$\displaystyle 2\sin \left ( x+y \right )=1+\sin \left ( x+y \right )$$
$$\displaystyle \sin \left ( x+y \right )=1\Rightarrow \cos(x+y)=0\Rightarrow x+y=\cfrac{\pi}{2}$$
Also the point lies on the given curve,
 $$y=0\Rightarrow x= \dfrac{\pi }{2}$$
Thus the point is, $$\displaystyle \left ( \frac{\pi }{2},0 \right )$$
Hence, option 'A' is correct.

$$\displaystyle  y=b\times e^{-\frac{x}{a}}$$

$$y=4x^3+3ax^3+6x^2+5$$
For given situation curve must be concave so,  $$\displaystyle\frac{d^2y}{dx^2}\,>\,0$$
$$y'=4x^3+9ax^2+12x$$
$$y''=12x^2+18ax+12\,>\,0=6(2x^2+3ax+2)\,>\,0$$
$$D\,<\,0\;\;\Rightarrow\;\;9a^2-16\,<\,0$$
$$\Rightarrow |a|\,<\,\displaystyle\frac{4}{3}$$

Let point be $$\displaystyle \left ( x_{1},y_{1} \right )$$
$$y = \displaystyle x^{3}+5$$
$$\Rightarrow \displaystyle \dfrac{dy}{dx}=3x^{2}$$
Now, the slope of tangent at this point is $$=\displaystyle \left ( \frac{dy}{dx} \right )_{x_{1}y_{1}}=3x_{1}^{2}$$
It is $$\displaystyle \perp $$ to $$x + 3y - 2 = 0$$
So $$\displaystyle 3x_{1}^{2}\times -\frac{1}{3}=-1\Rightarrow x_1=\pm 1 $$
Thus,
at $$\displaystyle x_{1}=1$$,             $$\Rightarrow$$                $$\displaystyle y_{1}=6$$
and at $$\displaystyle x_{1}=-1$$      $$\Rightarrow$$              $$\displaystyle y_{1}=4$$
Thus, the points are $$(1, 6)$$ and $$(-1, 4)$$
Hence, option 'D' is correct.

$$\displaystyle y^{2}=4a\left ( x+a\sin \frac{x}{a} \right )$$
$$\Rightarrow \displaystyle 2y\frac{dy}{dx}=4a\left ( 1+a\cos \frac{x}{a}\times \frac{1}{a} \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{2a}{y}\left ( 1+\cos \frac{x}{a} \right )$$
Now for tangent to be parallel to x-axis,
$$\displaystyle \frac{dy}{dx}=0=\frac{2a}{y}\left ( 1+\cos \frac{x}{a} \right )$$
$$\Rightarrow \displaystyle \frac{x}{a}=\pi $$
$$\Rightarrow \displaystyle x=a\pi $$
$$\therefore \displaystyle y^{2}=4a\left ( a\pi +a\times 0 \right )$$
$$\Rightarrow \displaystyle y^{2}=4a^{2}\pi =4ax$$
Clearly, this point lie on a parabola.
Hence, option 'B' is correct.

$$ln(x+y)=yln(x)$$

Given equation of curve $$x^3-y^3+x^2y-yx^2+3x-2y=0$$
Differentiating we get
$$\displaystyle 3{ x }^{ 2 }-3{ y }^{ 2 }\frac { dy }{ dx } +x^{ 2 }\frac { dy }{ dx } +2xy-2xy-x^{ 2 }\frac { dy }{ dx } +3-2\frac { dy }{ dx } =0$$

$$ (3{ y }^{ 2 }+2)\dfrac { dy }{ dx } =3{ x }^{ 2 }+3$$

$$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 3{ x }^{ 2 }+3 }{ (3{ y }^{ 2 }+2) } $$    

Slope of tangent to curve at (0,0) is $$m_1=\dfrac{3}{2}$$

Given equation of other curve $$x^5-y^4+2x+3y=0$$ 
On differentiation ,
$$\displaystyle 5{ x }^{ 4 }-4{ y }^{ 3 }\frac { dy }{ dx } +2+3\frac { dy }{ dx } =0$$

$$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 5{ x }^{ 4 }+2 }{ (4{ y }^{ 3 }-3) } $$
Slope of tangent to the curve at (0,0) is $$m_2=-\dfrac{2}{3}$$

Here, $$m_1m_2=-1$$
Hence, the lines are perpendicular.

Let the point on the curve $$\displaystyle 9y^{2}=x^{3}$$ be $$P\left ( x_{1},y_{1} \right ) $$
differentiating the curve w.r.t $$x$$
$$\displaystyle 2\times 9\times y\times \frac{dy}{dx}=3x^{2}$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{x^{2}}{2\times 3y}$$
$$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( x_{1},y_{1} \right )}=\frac{x_{1}^{2}}{2\times 3y_{1}}$$
Thus slope of Normal at P  $$=\displaystyle -\frac{dy}{dx}=\frac{-3y_{1}\times 2}{x_{1}^{2}}$$
Given normal makes equal intercept
$$\Rightarrow \displaystyle \frac{-3y_{1}\times 2}{x_{1}^{2}}=\pm 1$$
$$\Rightarrow \displaystyle \frac{-3y_{1}\times 2}{x_{1}^{2}}=1$$
$$\Rightarrow \displaystyle -3y_{1}\times 2=x_{1}^{2}$$ so $$9y_1^2\times 4=x_{1}^{4}\Rightarrow x_1^3 \times 4 = x_1^4$$
So $$\displaystyle x_{1}=4,\: \: \: y=-\frac{8}{3}$$
$$\Rightarrow \displaystyle \frac{-3y_{1}\times 2}{x_{1}^{2}}=-1 $$
$$\Rightarrow \displaystyle 2\times 3y_{1}=x_{1}^{2}$$                           ..........(1)
Also point lies point on the curve $$\displaystyle 9y^{2}=x^{3}$$
$$\Rightarrow \displaystyle 9y_{1}^{2}=x_{1}^{3}$$
$$\Rightarrow \displaystyle 9\times \frac{x_{1}^{4}}{36}=x_{1}^{3}$$
$$\Rightarrow \displaystyle x_{1}=4$$
$$\Rightarrow \displaystyle x_{1}=4,\: \: \: y_{1}=\frac{16}{6}=\frac{8}{3}$$
$$\Rightarrow \displaystyle \left ( 4,\frac{8}{3} \right )$$
Finally the points are $$\displaystyle \left ( 4,\frac{8}{3} \right )\: \: \: or\: \: \: \left ( 4,\frac{-8}{3} \right )$$
Hence, option 'A' is correct.

Given equation of curve is
$$y=b\ln x-x$$
Also, given $$x-y=0$$ is tangent to the curve.
So, let $$P(x_1,x_1)$$ be the point of tangency.
$$x_1=b\ln {x_1}-{x_1}$$
$$2x_1= b\ln {x_1}$$   ....(1)

Slope of tangent to the curve at P$$=\dfrac{b}{x_1}-1$$
Slope of given tangent $$=1$$
$$\Rightarrow \dfrac{b}{x_1}-1=1$$
$$\Rightarrow x_1=\dfrac{b}{2}$$
Substitute this value in (1), we get
$$b=b\ln {\dfrac{b}{2}}$$
$$\Rightarrow \ln {\dfrac{b}{2}}=1$$
$$\Rightarrow \dfrac{b}{2}=e$$
$$\Rightarrow b=2e$$
Since, $$2< e<3$$
$$\Rightarrow 4<b<6$$

$$\displaystyle { x }^{ 2m }{ y }^{ \tfrac { n }{ 2 }  }={ a }^{ \tfrac { 4m+n }{ 2 }  }$$
$$\displaystyle 2m\ln x +\frac { n }{ 2 } \ln y=\frac { 4m+n }{ 2 }\ln a$$
$$\displaystyle \frac { \left( 2m \right)  }{ x } +\left( \frac { n }{ 2 }  \right) \frac { 1 }{ y } \frac { dy }{ dx } =0$$
$$\displaystyle \frac { dy }{ dx } =\left( \frac { -2m }{ x }  \right) \frac { 2y }{ n } $$
Let $$P$$ be the point $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) $$
So equation of tangent is
$$\displaystyle y-{ y }_{ 1 }=\left( \frac { -4m }{ n } .\frac { { y }_{ 1 } }{ { x }_{ 1 } }  \right) \left( x-{ x }_{ 1 } \right) $$.......(i)
Let Tangents meets the axes at $$A$$ and $$B$$ respectively
$$\displaystyle A=\left( \frac { 4m+n }{ 4m } { x }_{ 1 },0 \right) \quad B=\left( 0,\frac { 4m+n }{ n } .{ y }_{ 1 } \right) $$
Let $$P$$ divides $$AB$$ in ratio $$\displaystyle \mu :1$$
$$\displaystyle \therefore \quad { x }_{ 1 }=\frac { \mu .0+1.\left( \dfrac { 4m+n }{ 4m }  \right) { x }_{ 1 } }{ \mu +1 } $$