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
\(\frac {x^2}{7}\) + \(\frac {y^2}{2}\) = 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 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 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 x² + 3y² = 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 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:
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 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:
Normal equation of the parabola y² = 4ax at t₁ be
y + xt₁ = at₁³ + 2at₁
It meets the point t₂ {at₂², 2at₂)
2 at₂ + at₂²t₁ = at₁³ + 2 at₁
at₂²t₁ – at₁³ = 2at₁ – 2at₂
at₁(t₂² – t₁²) = -2a(t₂ – t₁)
at₁(t₂ + t₁)(t₂ – t₁) = -2a(t₂ – t₁)
t₂ + t₁ = \(\frac {-2}{t_1}\)
t₂ = \(\frac {-2}{t_1}\) – t₁
t₂ = -(\(\frac {2}{t_1}\) + t₁)

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.