Tangents and its Equations Test 56

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Tangents and its Equations Test 56

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Question 1
1 / -0

If the curves $$\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{4} = 1$$ and $$y^{3} = 16x$$ intersect at right angles, then $$a^{2}$$ is equal to

A
$$5/3$$

B
$$4/3$$

C
$$6/11$$

D
$$3/2$$

Solution

$$y^{3}=16x$$
$$3y^{2}\dfrac{dy}{dx}=16$$
$$\dfrac{dy}{dx}=m_{1}=\dfrac{16}{3y^{2}}$$

$$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1$$
$$\dfrac{2x}{a^{2}}+\dfrac{2y}{4}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=\dfrac{-x}{a^{2}}\times \dfrac{4}{y}=m_{2}$$

$$m_{1}.m_{2}=-1$$
$$\dfrac{16}{3y^{2}}\left(\dfrac{-4x}{a^{2}y}\right)=-1$$
$$\dfrac{16\times 4\times x}{3a^{2}y^{3}}=1\Rightarrow 64x=3a^{2}y^{3}$$
$$a^{2}=\dfrac{64 x}{3y^{3}}$$ Now $$y^{3}=16 x$$
$$a^{2}=\dfrac{64 x}{3\times 16 x}\Rightarrow a^{2}=\dfrac{4}{3}$$
Question 2
1 / -0

If the slope of one of the lines given by $${a^2}{x^2} + 2hxy+by^2 = 0$$ be three times of the other , then h is equal to

A
$$2\sqrt 3 ab$$

B
$$-2\sqrt 3 ab$$

C
$$\frac{2}{{\sqrt 3 }}ab$$

D
$$-\frac{2}{{\sqrt 3 }}ab$$

Solution
Question 3
1 / -0

The equation of normal to the curve $$y=log^x_e$$ at the point $$P(1, 0)$$ is ___________.

A
$$2x+y=2$$

B
$$x-2y=1$$

C
$$x-y=1$$

D
$$x+y=1$$

Solution

To find the equation for the normal to the curve $$y = log_e x$$ at the point $$P(1,0)$$, we find the slope of said normal and then use the slope-point form of the equation of a line.
We know that the normal and the tangent of a curve at a point on the curve are perpendicular. So,
$$\text{Slope of the normal} = \frac{-1}{\text{Slope of the tangent}} = \frac{-1}{y'(1)}$$.
Now, $$y'(x) = \frac{1}{x}$$. So, $$y'(1) = 1$$.
Therefore, the slope of the normal to the curve $$y = log_e x$$ at the point $$P(1, 0)$$ is $$-1$$.
Noting that, the normal passes through the point $$P$$, we find its equation as:
$$y - 0 = -1 \times(x - 1)$$.
That is, $$x + y = 1$$.
Question 4
1 / -0

If the line joining the point (0, 3 ) and (5, -2) is a tangent to the curve $$y= \dfrac{c}{x+1}$$, then the value of c is

A
1

B
-2

C
4

D
None of these

Solution
Question 5
1 / -0

$$f(x) = \left\{\begin{matrix} -x^2, & \text{for} \ x < 0 \\ x^2 + 8, & \text{for} \ x \ge 0 \end{matrix}\right.$$
Let . Then x-intercept of the line, thet is , the tangent to the graph of f(x) is

A
zero

B
-1

C
-2

D
-4

Solution
Question 6
1 / -0

If the tangent at any point on the curve $$x+y^4=a$$ cuts off intercepts p and q on the co-ordinate axes, the value of $$p^{-4/3}+q^{-4/3}$$ is

A
$$a^{-4/3}$$

B
$$a^{-1/2}$$

C
$$a^{1/3}$$

D
None of these

Solution
Question 7
1 / -0

Let N be the set of positive integers. For all $$n \in N$$, let
$$f_n = (n + 1)^{1/3} - n^{1/3}$$ and $$A = \left\{n \in N : f_{n +1} < \dfrac{1}{3(n + 1)^{2/3}} < f_n \right\}$$
Then

A
A = N

B
A is a finite set

C
the complement of A in N is nonempty, but finite

D
A and its complement in N are both infinite

Solution
Question 8
1 / -0

At any point on the curve $$2x^{2}y^{2}-x^{4}=c$$, the mean proportional between the abscissa and the difference between the abscissa and the sub-normal drawn to the curve at the same point is equal to

A
ordinate

B
radius vector

C
x-intercept of tangent

D
sub-tangent

Solution
Question 9
1 / -0

Given $$g(x)= \dfrac{x+2}{x-1}$$ and the line 3x + y -10 =0, then the line is

A
tangent to g(x)

B
normal to g(x)

C
chord of g(x)

D
none of these

Solution
Question 10
1 / -0

The angle formed bt the positive y-axis and the tangent to $$y = x^{2}+4x-17$$ at $$(5/2, -3/4)$$ is

A
$$\tan ^{-1}(9)$$

B
$$\dfrac{\pi }{2}\tan ^{-1}(9)$$

C
$$\dfrac{\pi }{2}\tan ^{-1}(9)$$

D
None of these

Solution