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

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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 2x² + 7y² = 14 .
Solution:
2x² + 7y² = 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 a² = 7 and b² = 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 = 7m² + 2
25m² + 4 – 20m – 7m² – 2 = 0
18m² – 20m + 2 = 0
(÷ by 2) ⇒ 9m² – 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 a² = 16 and b² = 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 x² + 3y² = 12. Also find the coordinates of the point of contact.
Solution:
The given ellipse is x² + 3y² = 12
(÷ by 12) ⇒ \(\frac{x^{2}}{12}+\frac{y^{2}}{4}\) = 1
(ie.,) Here a² = 12 and b² = 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 c² = a²m² + b²
LHS = c² = 4² = 16
RHS: a²m² + b² = 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 y² = 16x perpendicular to 2x + 2y + 3 =0
Solution:
y² =16x
Comparing this equation with y² = 4ax
we get 4a = 16 ⇒ a = 4
The equation of tangent to the parabola y² = 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 y² = 8x. (Hint: use parametric form)
Solution:
y² = 8x .
Comparing this equation with y² = 4ax
we get 4a = 8 ⇒ a = 2
Now, the parametric form for y² = 4ax is x = at², 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 y² = 4 ax at (x₁, y₁) is yy₁ = 2a(x + x₁)
Here (x₁, y₁) = (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 y² = 4ax at ‘t’ is
yt = x + at²
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 12x² – 9y² = 108 at θ = \(\frac{\pi}{3}\) .
(Hint: Use parametric form)
Solution:
12x² – 9y² = 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 ‘t₁‘ and t₂’ on the parabola y² = 4ax is [at₁ t₂, a (t₁ + t₂)].
Solution:
The equation of tangent to y² = 4ax at ‘t’ is given by yt = x + at²
So equation of tangent at ‘t₁‘ is yt₁ = x + at₁²
and equation of tangent at ‘t₂‘ is yt₂ = x + at₂²
To find the point of intersection we have to solve the two equations

Question 8.
If the normal at the point ‘t₁‘ on the parabola y² = 4ax meets the parabola again at the point ‘t₂‘, then prove that t₂ = \(-\left(t_{1}+\frac{2}{t_{1}}\right)\)
Solution:
Equation of normal to y² =4 at’ t’ is y + xt = 2at + at³.
So equation of normal at ‘t₁’ is y + xt₁ = 2at₁ + at₁³
The normal meets the parabola y² = 4ax at ‘t₂’ (ie.,) at (at₂², 2at₂)
⇒ 2at₂ + at₁t₂² = 2at₁ + at₁³
So 2a(t₂ – t₁) = at₁³ – at₁t₂²
⇒ 2a(t₂ – t₁) = at₁(t₁² – t₂²)
⇒ 2(t₂ – t₁) = t₁(t₁ + t₂)(t₁ – t₂)
⇒ 2= -t₁(t₁ + t₂)
⇒ t₁ + t₂ = \(\frac{-2}{t_{1}}\)
⇒ t₂ = \(-t_{1}-\frac{2}{t_{1}}=-\left(t_{1}+\frac{2}{t_{1}}\right)\)

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 : x² + 2x – 4y + 4 = 0 at (0, 1)
Solution:

Question 2.
Find the equations of the tangent and normal to the parabola y² = 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 y² = 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 4x² – y² = 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 y² = 4x.
Solution:
y² = 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 2x² – 3y² = 6
Solution:
2x² – 3y² = 6

Question 7.
Prove that the line 5x + 12y = 9 touches the hyperbola x² – 9y² = 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 x² + 3y² = 12. Find the co-ordinates of the point of contact.
Solution:
The given ellipse is x² + 3y² = 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: a² m² + b = 12(1) + 4 = 16
LHS: RHS ⇒ the given lines is a tangent to the ellipse.