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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4

Question 1.

Find the equations of the two tangents that can be drawn from (5, 2) to the ellipse 2x^{2} + 7y^{2} = 14 .

Solution:

2x² + 7y² = 14

\(\frac {x^2}{7}\) + \(\frac {y^2}{2}\) = 1

y = mx + \(\sqrt { a^2m^2+b^2 }\)

a² = 7, b² = 2

It passes through (5, 2)

2 = 5 m + \(\sqrt { 7m^2 + 2 }\)

2 – 5 m = \(\sqrt { 7m^2 + 2 }\)

(2 – 5 m)² = (7m² + 2)

4 + 25m² – 20m = 7m² + 2

18m² – 20m + 2 = 0

9m² – 10m + 1 = 0

(9m – 1) (m – 1) = 0

m = 1 (or) m = \(\frac {1}{9}\)

Equation of tangent

(i) m= 1 ⇒ (y – 2) = 1(x- 5)

x – y – 3 = 0

(ii) m = \(\frac {1}{9}\) ⇒ y – 2 = \(\frac {1}{9}\) (x – 5)

9y – 18 = x – 5

x – 9y + 13 = 0

Equation of tangent be x – y – 3 = 0 and x – 9y + 13 = 0.

Question 2.

Find the equations of tangents to the hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{64}\) = 1 which are parallel to 10x – 3y + 9 = 0.

Solution:

\(\frac{x^{2}}{16}-\frac{y^{2}}{64}\) = 1

Here a^{2} = 16 and b^{2} = 64

The equation of tangents will be of the form y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

(i.e.,) y = mx ± \(\sqrt{16 m^{2}-64}\)

Where ‘m’ is the slope of the tangent.

Here we are given that the tangents are parallel to 10x – 3y + 9 = 0

So the slope of tangents will be equal to the slope of the given line

Question 3.

Show that the line x – y + 4 = 0 is a tangent to the ellipse x^{2} + 3y^{2} = 12. Also find the coordinates of the point of contact.

Solution:

Equation of the ellipse

x² + 3y² = 12

÷ by 12 ⇒ \(\frac {x^2}{12}\) + \(\frac {y^2}{4}\) = 1

a² = 12 b² = 4

Equation of the line

x – y + 4 = 0

y = x + 4

m = 1, c = 4

Condition that the line y = mx + c to be tangent to the ellipse is

c² = a²m² + b²

4² = 12(1)² + 4

16 = 16

∴ The line is a tangent to the ellipse.

Point of contact = (\(\frac {-a^2m}{c}\), \(\frac {b^2}{c}\))

= (\(\frac {-12(1)}{4}\), \(\frac {4}{4}\))

= (-3, 1)

Question 4.

Find the equation of the tangent to the parabola y^{2} = 16x perpendicular to 2x + 2y + 3 =0

Solution:

y^{2} =16x

Comparing this equation with y^{2} = 4ax

we get 4a = 16 ⇒ a = 4

The equation of tangent to the parabola y^{2} = 16x will be of the form

So m = 1

⇒ The equation of tangent will be y = 1(x) + \(\frac{4}{1}\) (i.e.,) y = x + 4

(or) x – y + 4 = 0

Question 5.

Find the equation of the tangent at t = 2 to the parabola y^{2} = 8x. (Hint: use parametric form)

Solution:

Equation of a tangent be

yt = x + at² Here t = 2

y² = 8x ∴ 4a = 8

a = 2

y(2) = x + 2(4)

2y = x + 8

x – 2y + 8 = 0

Question 6.

Find the equations of the tangent and normal to hyperbola 12x^{2} – 9y^{2} = 108 at θ = \(\frac{\pi}{3}\) .

(Hint: Use parametric form)

Solution:

12x^{2} – 9y^{2} = 108

Normal is a line perpendicular to tangent

So equation of normal will be of the form 3x + 4y + k = 0

The normal is drawn at (6, 6)

⇒ 18 + 24 + k = 0 ⇒ k = – 42

So equation of normal is 3x + 4y – 42 = 0

Question 7.

Prove that the point of intersection of the tangents at ‘t_{1}‘ and t_{2}’ on the parabola y^{2} = 4ax is [at_{1} t_{2}, a (t_{1} + t_{2})].

Solution:

The equation of tangent to y^{2} = 4ax at ‘t’ is given by yt = x + at^{2}

So equation of tangent at ‘t_{1}‘ is yt_{1} = x + at_{1}^{2}

and equation of tangent at ‘t_{2}‘ is yt_{2} = x + at_{2}^{2}

To find the point of intersection we have to solve the two equations

Question 8.

If the normal at the point ‘t_{1}‘ on the parabola y^{2} = 4ax meets the parabola again at the point ‘t_{2}‘, then prove that t_{2} = \(-\left(t_{1}+\frac{2}{t_{1}}\right)\)

Solution:

Normal equation of the parabola y² = 4ax at t_{1} be

y + xt_{1} = at_{1}³ + 2at_{1}

It meets the point t_{2} {at_{2}², 2at_{2})

2 at_{2} + at_{2}²t_{1} = at_{1}³ + 2 at_{1}

at_{2}²t_{1} – at_{1}³ = 2at_{1} – 2at_{2}

at_{1}(t_{2}² – t_{1}²) = -2a(t_{2} – t_{1})

at_{1}(t_{2} + t_{1})(t_{2} – t_{1}) = -2a(t_{2} – t_{1})

t_{2} + t_{1} = \(\frac {-2}{t_1}\)

t_{2} = \(\frac {-2}{t_1}\) – t_{1}

t_{2} = -(\(\frac {2}{t_1}\) + t_{1})

### Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4 Additional Questions

Question 1.

Find the equations of the tangent and normal to the parabolas: x^{2} + 2x – 4y + 4 = 0 at (0, 1)

Solution:

Question 2.

Find the equations of the tangent and normal to the parabola y^{2} = 8x at t = \(\frac{1}{2}\)

Solution:

∴ Equation of the tangent is 2x – y + 1 = 0 and equation of the normal is 2x + 4y – 9 = 0

Question 3.

Find the equations of the tangents: to the parabola y^{2} = 6x, parallel to 3x – 2y + 5 = 0

Solution:

∴ Equation of the tangent is 3x – 2y + 2 = 0

Question 4.

Find the equations of the tangents: to the parabola 4x^{2} – y^{2} = 64 Which are parallel to 10x – 3y + 9 = 0.

Solution:

Question 5.

Find the equation of the tangent from point (2, -3) to the parabola y^{2} = 4x.

Solution:

y^{2} = 4x

Equation of the tangent to the parabola will be of the form y = \(m x+\frac{1}{m}\)

The tangents pass through (2, -3)

The two tangents drawn from (2, -3) are x + y + 1 = 0, x + 2y + 4 = 0

Question 6.

Find the equation of the tangents from the point (2, -3) to the parabola 2x^{2} – 3y^{2} = 6

Solution:

2x^{2} – 3y^{2} = 6

Question 7.

Prove that the line 5x + 12y = 9 touches the hyperbola x^{2} – 9y^{2} = 9 and find the point of contact.

Solution:

(1) = (2) ⇒ The given line is a tangent to the curve i.e., the given line touches the curve. To find the point of contact we have to solve the line and the curve.

The given line 5x + 12y = 9

Question 8.

Show that the line x – y + 4 = 0 is a tangent to the ellipse x^{2} + 3y^{2} = 12. Find the co-ordinates of the point of contact.

Solution:

The given ellipse is x^{2} + 3y^{2} = 12

Given line is x – y + 4 = 0 ⇒ y = x + 4

Here, m = 1 and c = 4

The condition for the line y = mx + c to be a tangent to the ellipse

RHS: a^{2} m^{2} + b = 12(1) + 4 = 16

LHS: RHS ⇒ the given lines is a tangent to the ellipse.