Tangents and its Equations Test 12

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Tangents and its Equations Test 12

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Weekly Quiz Competition

Question 1
1 / -0

At origin the curve $$y^{2}=x^{3}+x$$

A
Touches the x-axis

B
Touches the y-axis

C
Bisects the angle between the axes

D
touches both the axes

Solution

$$\dfrac{dy}{dx}=\dfrac{3x^{2}+1}{2y}$$

At origin $$x=0,y=0$$

$$\tan\ \theta=\infty$$

$$\theta =\dfrac{\pi}{2}$$

$$\therefore$$ Curve touches the y-axis
Question 2
1 / -0

Observe the following statements
I: If $$p$$ and $$q$$ are the lengths of perpendiculars from the origin on the tangent and normal at any point on the curve $$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$$ then $$4p^{2}+q^{2}=1$$.
II: If the tangent at any point $$P$$ on the curve $$x^{3}.y^{2}=a^{5}$$ cuts the coordinate axes at $$A$$ and $$B$$ then $$AP : PB = 3 : 2$$

A
only I

B
only II

C
both I and II

D
neither I nor II

Solution

$$x=a\ \cos^{3}\theta$$
$$y=a\ \sin^{3}\theta$$
$$y'=-\tan\ \theta$$
$$(y-sin^{3}\theta)=-\tan\ \theta(x-\cos^{3}\theta)$$
$$y+n\ \tan\ \theta-\tan\ \theta\ \cos^{3}\theta-\sin^{3}\theta=0$$
Perpendicular from origin
$$\left | \dfrac{\tan\ \theta\ \cos^{3}\theta+\sin^{3}\theta}{\sqrt{1+\tan^{2}\theta}} \right |=p$$
$$p_{2}\ \ \sin\ \theta\ \cos\ \theta$$
$$p=\dfrac{\sin^{2}\theta}{2}$$
$$(y-\sin 3\theta)=\frac{1}{\tan\ \theta}(x-\cos3\theta)$$
$$y\ \tan\ \theta-\sin3\theta\ \tan\ \theta-x+\cos3\ \theta=0$$
Perpendicular frm the origin
$$\left | \dfrac{\cos3\theta-\sin3\theta\ \tan\ \theta}{\sqrt{1+\tan^{2}\theta}} \right |=q$$
$$q=\cos2\theta$$
$$4p^{2}+q^{2}=1$$
Question 3
1 / -0

Observe the following statements for the curve $$x
= at^{3}$$, $$y = at^{4}$$ at $$t = 1$$.
I : The equation of the tangent to the curve is $$4x-3y- a = 0$$
II : The equation of the normal to the curve is $$3x +4y- 7a = 0$$
III: Angle between tangent and normal at any point on the curve is $$\displaystyle \frac{\pi}{2}$$
Which of the above statements are correct.

A
I and II

B
II and III

C
I and III

D
I, II, III

Solution

$$\dfrac{dy}{dx}=\dfrac{4at^{3}}{3at^{2}}$$
$$\dfrac{dy}{dx}=\dfrac{4t}{3}$$
$$\dfrac{dy}{dx}|_{t=1}=\dfrac{4}{3}$$
$$(y-a)=\dfrac{4}{3}(x-a)$$ (tangent)
$$4x-3y-a=0$$
$$(y-a)=-\dfrac{3}{4}(x-a)$$ (Normal)
$$4y+3x-7a=0$$
Tangent and normal are always at an angle of $${\pi}/{2}$$ to each other
Question 4
1 / -0

Assertion (A): The normal to the curve $$ay^{2}=x^{3}(a\neq 0, x\neq 0)$$ at a point $$(x, y)$$ on it makes equal intercepts on the axes, then $$\displaystyle x=\frac{4a}{9}$$.
Reason (R): The normal at $$(x_{1}, y_{1})$$ on the curve $$y=f(x)$$ makes equal intercepts on the coordinate axes, then $$\left.\displaystyle \frac{dy} {dx}\right|_{(x_{1},y_{1})}=1$$

A
Both (A) and (R) are true and (R) is the correct explanation for (A).

B
Both (A) and (R) are true but (R) is not the correct explanation for (A).

C
(A) is true but (R) is false.

D
(A) is false but (R) is true.

Solution

$$ay^{2}=x^{3}$$
$$2ay\dfrac{dy}{dx}=3x^{2}\Rightarrow\dfrac{dy}{dx}=\dfrac{3x^{2}}{2ay}$$
Now, slope of normal
$$= -\dfrac{dx}{dy}=-\dfrac{2ay}{3x^{2}}=-\dfrac{2a\times x^{\frac{3}{2}}}{3x^{2}\sqrt{a}}=-\dfrac{2\sqrt{a}}{3\sqrt{x}}$$
For the normal to make equal intercepts on the axes, the slope has to be $$\pm1$$.
Hence, $$\dfrac{-2\sqrt{a}}{3\sqrt{x}}=\pm1$$
On squaring both sides,
$$\dfrac{4a}{9x}=1\Rightarrow x=\dfrac{4a}{9}$$.
Question 5
1 / -0

Assertion(A): If the tangent at any point $$P$$ on the curve $$xy = a^{2}$$ meets the axes at $$A$$ and $$B$$ then $$AP : PB = 1 : 1$$
Reason(R): The tangent at $$P(x, y)$$ on the curve $$X^{m}.Y^{n}=a^{m+n}$$ meets the axes at $$A$$ and $$B$$. Then the ratio of $$P$$ divides $$\overline{AB}$$ is $$n : m$$.

A
Both A and R are true R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

$$y=\dfrac{a^{2}}{x}$$
$$y'=-\dfrac{a^{2}}{x^{2}}$$
$$y'=-\dfrac{a^{2}}{t^{2}}$$
$$\left ( y-\dfrac{a^{2}}{t} \right )=\dfrac{-a^{2}}{t^{2}}(n-t)$$
$$B=(2+10)$$
$$A=\left ( 0,\ \dfrac{2a^{2}}{t} \right )$$
$$\dfrac{AP}{BP}=1$$
Question 6
1 / -0

The point of intersection of the tangents drawn to the curve $$x^{2}y=1-y$$ at the points where it is met by the curve $$xy=1-y$$ is given by

A
$$(0, -1)$$

B
$$(1, 1)$$

C
$$(0, 1)$$

D
$$(1, \infty)$$

Solution

$$x^{2}y=xy$$
$$y=1,\ x=0$$
$$y=^{1}/_{2},\ x=1$$
$$2x\dfrac{dx}{dy}=-\dfrac{-1}{y^{2}}$$
$$\dfrac{dy}{dx}=-2xy^{2}$$
$$(x=0)$$
$$\dfrac{dy}{dx}=0$$
$$(y-1)=x-0$$
$$y=1$$ ------- (1)
$$\left ( y-\dfrac{1}{2} \right )=\dfrac{-1}{2}(x-1)$$ -----(2)
$$x=0$$ (By (1) & (2))
$$(0, 1)$$
Question 7
1 / -0

The number of tangents to the curve $$x^{3/2}+y^{3/2}=a^{3/2}$$, where the tangents are equally inclined to the axes, is

A
$$2$$

B
$$1$$

C
$$0$$

D
$$4$$

Solution

$$\dfrac{dy}{dx}=-1$$ (where x is equally inclined)
$$\dfrac{3}{2}x^{^{1}/_{2}}+^{3}/_{2}y^{^{1}/_{2}}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=-\sqrt{\dfrac{x}{y}}=1$$
$$-\sqrt{\dfrac{x}{y}}=-1$$
$$\sqrt{x}=\sqrt{y}$$
$$x=y$$
$$x^{^{3}/_{2}}+x^{^{3}/_{2}}=a^{^{3}/_{2}}$$
$$2x^{^{3}/_{2}}=a^{^{3}/_{2}}$$
$$x=\dfrac{a}{2^{^{2}/_{3}}}$$
$$y=\dfrac{a}{2^{^{2}/_{3}}}$$
$$\therefore$$ only one tangent.
Question 8
1 / -0

The points on the hyperbola $$x^{2}-y^{2}=2$$ closest to the point (0, 1) are

A
$$(\displaystyle \pm\frac{3}{2},\frac{1}{2})$$

B
$$(\displaystyle \frac{1}{2},\pm\frac{3}{2})$$

C
$$(\displaystyle \frac{1}{2},\frac{1}{2})$$

D
$$(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})$$

Solution

Let point be $$(\sqrt 2 sec\theta,\sqrt 2 tan\theta)$$ closest to (0, 1)
So distance $$=\sqrt {2 sec^2\theta+(\sqrt 2 tan\theta-1)^2}$$
$$=\sqrt {2 sec^2\theta+2tan^2\theta-2\sqrt 2 tan\theta+1}$$
$$=\sqrt {4 tan^2\theta-2\sqrt 2 tan\theta+3}$$
$$=2\sqrt {tan^2\theta-\frac {tan\theta}{\sqrt 2}+\frac {3}{4}}$$
$$=2\sqrt {(tan\theta-\frac {1}{2\sqrt 2})^2+\frac {3}{4}-\frac {1}{8}}$$
So distance is min when $$tan\theta=\frac {1}{2\sqrt 2}$$
$$sec\theta=\pm \frac {3}{2\sqrt 2}$$
Question 9
1 / -0

If the circle $$x^2 + y^2 + 2gx + 2fy + c =0$$ is touched by y = x at P in the first quadrant, such that $$OP = 6 \sqrt2$$, then the value of $$c$$ is

A
36

B
144

C
72

D
None of these

Solution

Let the point of contact be $$x_{1},y_{1}$$
However it lies on the line $$y=x$$
Hence
$$x_{1}=y_{1}$$
Applying distance formula, we get
$$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$$
$$2x_{1}^2=72$$
$$x_{1}=6$$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
$$2x+2yy'+2g+2fy'=0$$
Now $$y'=1$$ since $$y'$$ is the slope of the line $$y=x$$.
By substituting, we get
$$x+y=-(g+f)$$
Now $$x_{1}=y_{1}=6$$
Hence
$$12=-(g+f)$$ ...(i)
Substituting $$x=y=6$$ in the equation of the circle, we get
$$36+36+2(6)(g+f)+c=0$$
$$72+12(-12)+c=0$$
$$c=144-72$$
$$c=72$$
Question 10
1 / -0

lf the curve $$y=px^{2}+qx+r$$ passes through the point (1, 2) and the line $$y=x$$ touches it at the origin, then the values of $$p,\ q$$ and $$r$$ are

A
$$p=1,q=-1,r=0$$

B
$$p=1,q=1,r=0$$

C
$$p=-1,q=1,r=0$$

D
$$p=1,q=2,r=3$$

Solution

$$2=p+q+r$$
$$r=0$$
$$\dfrac{dx}{dy}|_{x=0}=1$$
$$\dfrac{dy}{dx}=2px+q$$
$$q=1$$
$$2=p+1+0$$
$$p=1$$

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Tangents and its Equations Test 12

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Tangents and its Equations Test 12

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SHARING IS CARING

Question 1

Touches the x-axis

Touches the y-axis

Bisects the angle between the axes

touches both the axes

Solution

Question 2

only I

only II

both I and II

neither I nor II

Solution

Question 3

I and II

II and III

I and III

I, II, III

Solution

Question 4

Both (A) and (R) are true and (R) is the correct explanation for (A).

Both (A) and (R) are true but (R) is not the correct explanation for (A).

(A) is true but (R) is false.

(A) is false but (R) is true.

Solution

Question 5

Both A and R are true R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 6

$$(0, -1)$$

$$(1, 1)$$

$$(0, 1)$$

$$(1, \infty)$$

Solution

Question 7

$$2$$

$$1$$

$$0$$

$$4$$

Solution

Question 8

$$(\displaystyle \pm\frac{3}{2},\frac{1}{2})$$

$$(\displaystyle \frac{1}{2},\pm\frac{3}{2})$$

$$(\displaystyle \frac{1}{2},\frac{1}{2})$$

$$(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})$$

Solution

Question 9

36

144

72

None of these

Solution

Question 10

$$p=1,q=-1,r=0$$

$$p=1,q=1,r=0$$

$$p=-1,q=1,r=0$$

$$p=1,q=2,r=3$$

Solution

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