Tangents and its Equations Test 23

Let the point is $$P(a,b)$$
Now the given curve is $$y=2x^2-x+1$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=4x-1$$
Thus slope of tangent at P is $$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=4a-1$$
But given the tangent is parallel to line $$y=3x+4$$
$$\Rightarrow 4a-1=3\Rightarrow a=1$$
Also the point P lies on the given curve,
$$b=2a^2-a+1=2$$
Therefore, the point P is $$(1,2)$$
Hence, option 'B' is correct.

The slope of normal $$\displaystyle x=a\left ( \theta -\sin \theta

 \right );\: \: \: \: y=a\left ( 1-\cos \theta  \right )\: \: \: at\: \:

\: \theta =\dfrac{\pi }{2}$$
$$\displaystyle \left ( \dfrac{dx}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\left ( 1-\cos \theta  \right )=a $$
$$\displaystyle \left ( \dfrac{dy}{d\theta } \right )_{\theta =\dfrac{\pi }{2}}=a\sin \theta =a $$
$$\therefore \left ( \dfrac{dy}{dx} \right )=1$$
Therefore, slope of normal at the given point is, $$m= \left ( -\dfrac{dx}{dy} \right )=-1$$
Hence, option 'C' is correct.

$$\displaystyle y=\frac { x+9 }{ x+5 } =1+\frac { 4 }{ x+5 } $$

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.

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

$$y^2=4ax$$
$$\Rightarrow 2y\cfrac{dy}{dx}=4a\Rightarrow \cfrac{dy}{dx}=\cfrac{2a}{y}$$
Therefore, slope of normal to the given curve is, $$=\left(-\cfrac{dx}{dy}\right)_{(at^2,2at)}=-\cfrac{2at}{2a}=-t$$
Hence, option 'C' is correct.

Given  curve is $$xy + ax - by = 0$$
Slope of tangent at $$(1, 1)$$ is $$\displaystyle x.\frac{dy}{dx}+y+a-b.\frac{dy}{dx}=0 $$
$$\displaystyle

\left ( \frac{dy}{dx} \right )_{\left ( 1,1 \right )}=-\left [

\frac{\left ( a+y \right )}{x-b} \right ]_{\left ( 1,1 \right

)}=-\frac{\left ( a+1 \right )}{1-b} $$
$$\displaystyle \therefore -\frac{\left ( a+1 \right )}{1-b}=2 $$
So, $$2b - a = 3  $$                                        ..........(1)
$$\displaystyle \because (1, 1)$$ lies on curve $$xy + ax - by = 0$$
$$\displaystyle \therefore a-b=-1 $$         .........(2)
From (1) and (2), we get
$$a = 1, b = 2$$
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
$$\dfrac{12a-24}{16a-46}=\dfrac{1}{0}$$
$$\Rightarrow 16a-46=0$$
$$\Rightarrow a=\dfrac{23}{8}$$