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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2

Question 1.

Find the slope of the tangent to the curves at the respective given points.

(i) y = x^{4} + 2x^{2} – x at x = 1

(ii) x = a cos^{3} t, sin ^{3} t at t = \(\frac{\pi}{2}\)

Solution:

(i) y = x^{4} + 2x^{2} – x at x = 1

Question 2.

Find the point on the curve y = x^{2} – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.

Solution:

Let (x_{1}, y_{1}) be the required point.

Given tangent is similarly to the line

⇒ m_{1} = m_{2}

⇒ 2x_{1} – 5 = -3

⇒ 2x_{1} = -3 + 5 = 2

⇒ x_{1} = 1

Substituting x_{1} = 1 in the curve.

y_{1} = 1 – 5 + 4 = 0

So the required point is (1, 0)

Question 3.

Find the points on the curves = x^{3} – 6x^{2} + x + 3 where the normal is parallel to the line x + y = 1729.

Solution:

Let (x_{1},y_{1}) be the required point

y = x^{3} – 6x^{2} + x + 3

Substituting x_{1} values in the curve

when x_{1} = 4, y_{1} = 3

when x_{1} = 4, y_{1} = 4^{3} – 6(4)^{2} + 4 + 3 = 64 – 96 + 4 + 3 = -25

So the required points are (0, 3) and (4, -25)

Question 4.

Find the points on the curve y^{2} – 4xy = x^{2} + 5 for which the tangent is horizontal.

Solution:

Given that the tangent is horizontal (i.e) tangent is parallel to the x-axis

⇒ Equation of tangent will be of the form y = c

from (1) and (2) we get

Substituting x = -2y in the equation of the curve we get

y^{2} – 4(-2y)(y) = 4y^{2} + 5

⇒ y^{2} + 8y^{2} – 4y^{2} = 5

⇒ 5y^{2} = 5 ⇒ y^{2} = 1

⇒ y = ± 1

when y = 1, x = -2 and when y = -1, x = 2 So the points are (2, -1), (-2, 1)

Question 5.

Find the tangent and normal to the following curves at the given points on the curve.

Solution:

Now m = -2, (x_{1}, y_{1}) = (1, 0)

So equation of the tangent is

y – y_{1} = m(x – x_{1})

(i. e) y – 0 = -2(x – 1)

y = -2x + 2

2x+ y = 2

Slope of tangent = m = -2

Question 6.

Find the equations of the tangents to the curve y = 1 + x^{3} for which the tangent is orthogonal with the line x + 12y = 12.

Solution:

Substituting x_{1} values in the curve

when x_{1} = 2, y_{1} = 9; when x_{1} = -2, y_{1} = -1

So the points are (2, 9) and (-2, -7)

To find the equations of tangents:

Tangents are orthogonal to x + 12y = 12

So equations of tangents will be of the form 12x – y = k

The tangent passes through (2, 9) ⇒ 24 – 9 = k ⇒ k = 15 .

∴ Equation of tangent is 12x – y = 15

The tangent passes through (-2, -7) ⇒ 12 (-2) + 7 = k ⇒ -17

So equation of tangent is 12x – y = -17

Question 7.

Find the equations of the tangents to the curve y = \(\frac{x+1}{x-1}\)which are parallel to the line x + 2y = 6.

Solution:

Tangent is parallel the line x + 2y = 6

⇒ The slope of tangent = Slope of a line

when x_{1} = -1, y_{1} = 0; when x_{1} = 3, y_{1} = 2

So the points are (-1, 0) and (3, 2). The tangents are parallel to x + 2y = 6.

So the equation of tangents will be of the form x + 2y = k.

∴ Equation of tangent is x + 2y = -1

The tangent passes through (-1, 0) ⇒ -1 = k

The tangent passes through (3, 2) ⇒ 3 + 4 = k ⇒ k = 7

∴ Equation of tangent is x + 2y = 7

Question 8.

Find the equation of tangent and normal to the curve given by x = 7 cos t and y = 2sin t, t ∈ R at any point on the curve.

Solution:

(2 cos t)y – 4 sint cos t = (7 sin t)x – 49 sin t cos t

x(7 sin t) – y(2 cos t) = 45 sin t cos t

Question 9.

Find the angle between the rectangular hyperbola xy = 2 and the parabola x^{2} + 4y = 0.

Solution:

Solving the given two equations. To find the point of intersection:

Question 10.

Show that the two curves x^{2} – y^{2} = r^{2} and xy = c^{2} where c, r are constants, cut orthogonally.

Solution:

Let (x_{1}, y_{1}) be the point of intersection of the two curves

I Curve: x^{2} – y^{2} = r^{2}

Differentiating w.r.to x

II Curve: xy = c^{2}

Differentiating w.r.to x

### Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2 Additional Questions

Question 1.

Prove that the sum of the intercepts on the co-ordinate axes of any tangent to the curve x = a cos^{4} θ, y = a sin^{4} θ, 0 < θ < \(\frac{\pi}{2}\) is equal to a.

Solution:

Question 2.

Find the equations of a normal to y = x^{3} – 3x that is parallel to 2x + 18y – 9 = 0.

Solution:

Let (x_{1}, y,) be a point on the curve.

It is given that the normal is parallel to the line ⇒ m_{1 }= m_{2}

9y + 18 = -x – 2

x + 9y + 20 = 0

Question 3.

Prove that the curves 2x^{2} + 4y^{2} = 1 and 6x^{2} – 12y^{2} = 1 cut each other at right angles.

Solution:

Solving the given two equations,

Similarly it can be proved at the other points also.

Question 4.

Show that the equation of the normal to the curve x = a cos^{3} θ, y = a sin^{3} θ at ‘θ’ is x cos θ – y sin θ = a cos 2θ

Solution:

So, equation of the normal is

i.e. y sin θ – a sin^{4} θ = x cos θ – a cos^{4} θ

x cos θ – y sin θ = a cos^{4} θ – a sin^{4} θ

i.e.x cos θ – y sin θ = a[cos^{2} θ + sin^{2}θ][cos^{2}θ – sin^{2} θ]

= a[cos 2 θ]

So, the equation of the normai is x cos θ – y sin θ = a cos 2θ

Question 5.

If the curve y^{2} = x and xy = k are orthogonal, then prove that 8k^{2} = 1.

Solution:

y^{2} = x, xy = k

Solving the two equations, we get, (y^{2}) (y) = k

y^{3} = k