Application of Derivatives Test - 31

$$\displaystyle y^{2}=4a\left ( x+a\sin \frac{x}{a} \right )$$
$$\Rightarrow \displaystyle 2y\frac{dy}{dx}=4a\left ( 1+a\cos \frac{x}{a}\times \frac{1}{a} \right )$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{2a}{y}\left ( 1+\cos \frac{x}{a} \right )$$
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.

Given $$r = 7.5 cm $$ ,$$\dfrac {dr} {dt}=3.5 cm/sec$$
$$A = \pi r^2$$
$$\dfrac {dA}{dt}  =2\pi r \times \dfrac {dr}{dt}$$

$$\dfrac{dA}{dt}= 2  \pi \times  7.5 \times  3.5 $$

$$\dfrac {dA}{dt}=52.5 \pi  cm^2/sec $$

Let the point be $$P\displaystyle \left ( x_{1},y_{1} \right )$$
Now $$\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{a}$$
Differentiating w.r.t $$x$$
$$\displaystyle \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\times \frac{dy}{dx}=0 $$
$$\displaystyle \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$$
Thus, slope of normal at P is $$ =\displaystyle -( \frac{dx}{dy}  )_{( x_{1},y_{1} )}=-\sqrt{\frac{x_{1}}{y_{1}}} $$
If normal $$\displaystyle \perp $$ to x axis then $$\displaystyle \frac{dx}{dy}=\frac{a}{0} $$
$$\displaystyle \therefore  y_1 = a\Rightarrow  x_1 = 0$$
Therefore, required point is  $$(a, 0)$$
Hence, option 'B' is correct.

$$\displaystyle x^{3}-y^{2}=0 $$
$$\Rightarrow \displaystyle 3x^{2}-2y\left ( \frac{dy}{dx} \right )=0 $$
$$\Rightarrow \displaystyle 3x^{2}=2y\left ( \frac{dy}{dx} \right ) $$
$$\Rightarrow \displaystyle \frac{3x^{2}}{2y}=\frac{dy}{dx}$$
$$\Rightarrow \displaystyle \left ( \frac{dy}{dx} \right )_{\left ( m^{2},-m^{3} \right )}=\frac{3\times m^{4}}{-2\times m^{3}}$$
$$\therefore \displaystyle -\left ( \frac{dx}{dy} \right )=\frac{2}{3m}$$
Therefore, equation of normal at given point is,
$$\Rightarrow \displaystyle \left ( y+m^{3} \right )=\frac{2}{3m}\left ( x-m^{2} \right )$$
$$\Rightarrow \displaystyle 3my+3m^{4}=2x-2m^{2}$$
$$\Rightarrow \displaystyle 3my=2x-3m^{4}-2m^{2}$$
$$\Rightarrow \displaystyle y=\frac{2x}{3m}-m^{3}-\frac{2}{3}m ..(1)$$
but given equation of normal is, $$\displaystyle y=3mx-4m^{3} ..(2)$$
Thus comparing (1) and (2), we get
$$\displaystyle 3m=\frac{2}{3m}$$
$$\Rightarrow \displaystyle m^{2}=\frac{2}{9}$$
Hence, option 'D' is correct.

$$y = (x + 1)(x - 3) ..(1)$$

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.

The co-ordinates are $$x = \displaystyle a\left ( \theta +\sin \theta  \right );\: \: y=a\left ( 1-\cos \theta  \right )$$
Thus, $$\displaystyle \frac{dx}{d\theta }=a\left ( 1+\cos \theta  \right )\displaystyle =a\left ( 2\cos^{2}\frac{\theta }{2}\right )$$ and $$\displaystyle \frac{dy}{d\theta }=a\sin \theta =2a\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$$
$$\therefore \displaystyle \dfrac{dy}{dx}=\tan \dfrac{\theta }{2}$$
but, $$\displaystyle \frac{dy}{dx}=\tan \frac{\pi }{4}$$       (given)
$$\therefore 1 = \tan \dfrac{\theta }{2}$$
$$\Rightarrow \displaystyle \theta =\frac{\pi }{2}$$
So the point is, $$\displaystyle x=a\left ( \theta +\sin \theta  \right )$$
$$x = \displaystyle a\left ( \frac{\pi }{2}+1 \right )$$
$$y = a(1 - 0)=a$$
Therefore, the required point is $$\displaystyle \left ( a\left ( \frac{\pi }{2}+1 \right ),a \right )$$
Hence, option 'C' is correct.

Given equation of curve
$$y^2=ax^3-6x^2+b$$     
Since, the curve passes through $$(0,1)$$
$$\Rightarrow b=1$$

Since, the tangent is parallel to y-axis at $$x=2$$
$$\Rightarrow y=8a-23$$
So, point of tangency is $$(2,8a-23)$$

Slope of tangent to curve
$$2y\dfrac{dy}{dx}=3ax^2-12x$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{3ax^2-12x}{2y}$$
Slope of tangent to curve at $$(2,8a-23)$$ is $$\dfrac{12a-24}{16a-46}$$

Since, the tangent is parallel to y-axis
$$\dfrac{12a-24}{16a-46}=\dfrac{1}{0}$$
$$\Rightarrow 16a-46=0$$
$$\Rightarrow a=\dfrac{23}{8}$$

$$\displaystyle { x }^{ 2m }{ y }^{ \tfrac { n }{ 2 }  }={ a }^{ \tfrac { 4m+n }{ 2 }  }$$
$$\displaystyle 2m\ln x +\frac { n }{ 2 } \ln y=\frac { 4m+n }{ 2 }\ln a$$
$$\displaystyle \frac { \left( 2m \right)  }{ x } +\left( \frac { n }{ 2 }  \right) \frac { 1 }{ y } \frac { dy }{ dx } =0$$
$$\displaystyle \frac { dy }{ dx } =\left( \frac { -2m }{ x }  \right) \frac { 2y }{ n } $$
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 $$\displaystyle \mu :1$$
$$\displaystyle \therefore \quad { x }_{ 1 }=\frac { \mu .0+1.\left( \dfrac { 4m+n }{ 4m }  \right) { x }_{ 1 } }{ \mu +1 } $$

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