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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:

2x^{2} + 7y^{2} = 14

(÷ by 14) ⇒ \(\frac{x^{2}}{7}+\frac{y^{2}}{2}\) = 1

comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1

we get a^{2} = 7 and b^{2} = 2

The equation of tangent to the above ellipse will be of the form

y = mx + \(\sqrt{a^{2} m^{2}+b^{2}}\) ⇒ y = mx + \(\sqrt{7 m^{2}+2}\)

Here the tangents are drawn from the point (5, 2)

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

Squaring on both sides we get

(2 – 5m)^{2} = 7m^{2} + 2

25m^{2} + 4 – 20m – 7m^{2} – 2 = 0

18m^{2} – 20m + 2 = 0

(÷ by 2) ⇒ 9m^{2} – 10m + 1 = 0

(9m – 1) (m – 1) = 0

‘ m = 1 (or) m = 1/9

When m = 1, the equation of tangent is

y = x + 3 or x – y + 3 = 0

When m = \(\frac{1}{9}\) the equation of tangent is 9

y = \(=\frac{x}{9}+\sqrt{\frac{7}{81}+2}\) (i.e.) y = \(\frac{x}{9}+\frac{13}{9}\)

9y = x + 13 or 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 slope of tangents will be equatl to 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:

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

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

(ie.,) Here a^{2} = 12 and b^{2} = 4

The given line is x – y + 4 = 0

(ie.,) y = x + 4

Comparing this line with y = mx + c

We get m = 1 and c = 4

The condition for the line y = mx + c

To be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is c^{2} = a^{2}m^{2} + b^{2}

LHS = c^{2} = 4^{2} = 16

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

LHS = RHS The given line is a tangent to the ellipse. Also the point of contact is

\(\left(\frac{-a^{2} m}{c}, \frac{b^{2}}{c}\right)=\left[-\left(\frac{12(1)}{4}\right), \frac{4}{4}\right]\) (i.e.,) (-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:

y^{2} = 8x .

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

we get 4a = 8 ⇒ a = 2

Now, the parametric form for y^{2} = 4ax is x = at^{2}, y = 2at

Here a = 2 and t = 2

⇒ x = 2(2)^{2} = 8 and y = 2(2) (2) = 8

So the point is (8, 8)

Now eqution of tangent to y^{2} = 4 ax at (x_{1}, y_{1}) is yy_{1} = 2a(x + x_{1})

Here (x_{1}, y_{1}) = (8, 8) and a = 2

So equation of tangent is y(8) = 2(2) (x + 8)

(ie.,) 8y = 4 (x + 8)

(÷ by 4) ⇒ 2y = x + 8 ⇒ x – 2y + 8 = 0

Aliter

The equation of tangent to the parabola y^{2} = 4ax at ‘t’ is

yt = x + at^{2}

Here t = 2 and a = 2

So equation of tangent is

(i.e.,) y(2) = x + 2(2)^{2}

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:

Equation of normal to y^{2} =4 at’ t’ is y + xt = 2at + at^{3}.

So equation of normal at ‘t_{1}’ is y + xt_{1} = 2at_{1} + at_{1}^{3}

The normal meets the parabola y^{2} = 4ax at ‘t_{2}’ (ie.,) at (at_{2}^{2}, 2at_{2})

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

So 2a(t_{2} – t_{1}) = at_{1}^{3} – at_{1}t_{2}^{2}

⇒ 2a(t_{2} – t_{1}) = at_{1}(t_{1}^{2} – t_{2}^{2})

⇒ 2(t_{2} – t_{1}) = t_{1}(t_{1} + t_{2})(t_{1} – t_{2})

⇒ 2= -t_{1}(t_{1} + t_{2})

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

⇒ t_{2} = \(-t_{1}-\frac{2}{t_{1}}=-\left(t_{1}+\frac{2}{t_{1}}\right)\)

### 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.