Samacheer Kalvi Class 12 Maths Solutions Chapter 5 Two Dimensional Analytical Geometry II Exercise 5.4

Question 1. Find the equations of the two tangents that can be drawn from (5, 2) to the ellipse \( 2x^2 + 7y^2 = 14 \).
Answer: The equation of the ellipse is \( 2x^2 + 7y^2 = 14 \).
We can divide by 14 to get the standard form: \( \frac{2x^2}{14} + \frac{7y^2}{14} = 1 \)
\( \frac{x^2}{7} + \frac{y^2}{2} = 1 \)
Comparing this with the standard equation \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1 \), we find that \( a^2 = 7 \) and \( b^2 = 2 \).
The general equation for a tangent to an ellipse is \( y = mx + \sqrt{a^{2} m^{2}+b^{2}} \).
Substituting \( a^2 = 7 \) and \( b^2 = 2 \), we get: \( y = mx + \sqrt{7 m^{2}+2} \).
The tangents are drawn from the point (5, 2). So, we substitute x = 5 and y = 2 into the tangent equation:
\( 2 = 5m + \sqrt{7 m^{2}+2} \)
To solve for m, we isolate the square root term:
\( 2 – 5m = \sqrt{7 m^{2}+2} \)
Now, we square both sides to remove the square root:
\( (2 – 5m)^2 = ( \sqrt{7 m^{2}+2} )^2 \)
\( 4 - 20m + 25m^2 = 7m^2 + 2 \)
Rearranging the terms to form a quadratic equation:
\( 25m^2 - 7m^2 - 20m + 4 - 2 = 0 \)
\( 18m^2 - 20m + 2 = 0 \)
Divide the entire equation by 2 to simplify:
\( \implies 9m^2 - 10m + 1 = 0 \)
Factor this quadratic equation:
\( 9m^2 - 9m - m + 1 = 0 \)
\( 9m(m - 1) - 1(m - 1) = 0 \)
\( (9m - 1)(m - 1) = 0 \)
This gives us two possible values for m:
\( 9m - 1 = 0 \implies m = \frac{1}{9} \)
\( m - 1 = 0 \implies m = 1 \)

Now we find the equation of the tangent for each value of m using \( y = mx + c \), where \( c = \sqrt{7m^2+2} \).

Case 1: When \( m = 1 \)
\( y = 1x + \sqrt{7(1)^2+2} \)
\( y = x + \sqrt{7+2} \)
\( y = x + \sqrt{9} \)
\( y = x + 3 \)
Rearranging this equation into the general form:
\( x - y + 3 = 0 \)

Case 2: When \( m = \frac{1}{9} \)
\( y = \frac{1}{9}x + \sqrt{7(\frac{1}{9})^2+2} \)
\( y = \frac{x}{9} + \sqrt{\frac{7}{81}+2} \)
\( y = \frac{x}{9} + \sqrt{\frac{7+162}{81}} \)
\( y = \frac{x}{9} + \sqrt{\frac{169}{81}} \)
\( y = \frac{x}{9} + \frac{13}{9} \)
Multiply the entire equation by 9 to remove fractions:
\( 9y = x + 13 \)
Rearranging this equation into the general form:
\( x - 9y + 13 = 0 \)

Thus, the two equations of the tangents are \( x - y + 3 = 0 \) and \( x - 9y + 13 = 0 \). The process involves using the general tangent equation and then solving a quadratic equation for the slope 'm'.
In simple words: We found two straight lines that touch the ellipse at exactly one point, starting from a given outside point. We did this by using a special formula for tangents and solving for the slope of these lines.

🎯 Exam Tip: Remember to check for extraneous solutions after squaring both sides of an equation involving square roots. Always verify your 'm' values by substituting them back into the original equation before squaring.

Question 2. Find the equations of tangents to the hyperbola \( \frac {x^2}{16} – \frac {y^2}{64} = 1 \) which are parallel to \( 10x – 3y + 9 = 0 \).
Answer: The equation of the hyperbola is \( \frac {x^2}{16} – \frac {y^2}{64} = 1 \).
Comparing this with the standard hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we get \( a^2 = 16 \) and \( b^2 = 64 \).

The given line to which the tangents are parallel is \( 10x – 3y + 9 = 0 \).
To find the slope of this line, we rearrange it into the form \( y = mx + c \):
\( 3y = 10x + 9 \)
\( y = \frac{10}{3}x + \frac{9}{3} \)
So, the slope of this line is \( m = \frac{10}{3} \).
Since the required tangents are parallel to this line, their slope will also be \( m = \frac{10}{3} \).

The condition for a line \( y = mx + c \) to be a tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by \( c^2 = a^2m^2 – b^2 \).
Substitute the values: \( a^2 = 16 \), \( b^2 = 64 \), and \( m = \frac{10}{3} \).
\( c^2 = 16 \left( \frac{10}{3} \right)^2 – 64 \)
\( c^2 = 16 \left( \frac{100}{9} \right) – 64 \)
\( c^2 = \frac{1600}{9} – 64 \)
To subtract, find a common denominator:
\( c^2 = \frac{1600}{9} – \frac{64 \times 9}{9} \)
\( c^2 = \frac{1600 - 576}{9} \)
\( c^2 = \frac{1024}{9} \)
Now, take the square root to find c:
\( c = \pm \sqrt{\frac{1024}{9}} \)
\( c = \pm \frac{32}{3} \)

We have the slope \( m = \frac{10}{3} \) and two values for c: \( c = \frac{32}{3} \) and \( c = -\frac{32}{3} \).
The equations of the tangents are in the form \( y = mx + c \).

Tangent 1: \( y = \frac{10}{3}x + \frac{32}{3} \)
Multiply by 3 to clear the denominators:
\( 3y = 10x + 32 \)
Rearrange to general form:
\( 10x - 3y + 32 = 0 \)

Tangent 2: \( y = \frac{10}{3}x - \frac{32}{3} \)
Multiply by 3 to clear the denominators:
\( 3y = 10x - 32 \)
Rearrange to general form:
\( 10x - 3y - 32 = 0 \)

So, the equations of the two tangents parallel to the given line are \( 10x - 3y + 32 = 0 \) and \( 10x - 3y - 32 = 0 \). These lines have the same slope as the given line but different y-intercepts, touching the hyperbola at distinct points.
In simple words: We found two lines that are parallel to a given line and also touch the hyperbola. We used a special rule that connects the slope of the tangent, the hyperbola's numbers, and the point where it touches.

🎯 Exam Tip: Remember that parallel lines have the same slope. When dealing with hyperbolas, the sign in the tangent condition \( c^2 = a^2m^2 – b^2 \) is crucial and differs from the ellipse condition.

Question 3. Show that the line \( x - y + 4 = 0 \) is a tangent to the ellipse \( x^2 + 3y^2 = 12 \). Also find the co-ordinates of the point of contact.
Answer: The equation of the ellipse is \( x^2 + 3y^2 = 12 \).
To convert it to standard form, divide by 12:
\( \frac{x^2}{12} + \frac{3y^2}{12} = \frac{12}{12} \)
\( \frac{x^2}{12} + \frac{y^2}{4} = 1 \)
Comparing this with \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we find \( a^2 = 12 \) and \( b^2 = 4 \).

The given line is \( x - y + 4 = 0 \).
Rearrange this to the slope-intercept form \( y = mx + c \):
\( y = x + 4 \)
From this, we can see that the slope \( m = 1 \) and the y-intercept \( c = 4 \).

For a line \( y = mx + c \) to be a tangent to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the condition is \( c^2 = a^2m^2 + b^2 \).
Let's check if this condition holds true for our values:
Left Hand Side (LHS): \( c^2 = 4^2 = 16 \).
Right Hand Side (RHS): \( a^2m^2 + b^2 = 12(1)^2 + 4 \)
\( = 12(1) + 4 \)
\( = 12 + 4 \)
\( = 16 \).
Since LHS = RHS (\( 16 = 16 \)), the condition is satisfied. Therefore, the line \( x - y + 4 = 0 \) is indeed a tangent to the ellipse \( x^2 + 3y^2 = 12 \).

Now, to find the co-ordinates of the point of contact, we use the formula for the point of contact \( (x_1, y_1) \) for an ellipse:
\( (x_1, y_1) = \left( \frac{-a^2 m}{c}, \frac{b^2}{c} \right) \)
Substitute the values: \( a^2 = 12 \), \( b^2 = 4 \), \( m = 1 \), and \( c = 4 \).
\( x_1 = \frac{-12 \times 1}{4} = \frac{-12}{4} = -3 \)
\( y_1 = \frac{4}{4} = 1 \)
So, the co-ordinates of the point of contact are \( (-3, 1) \). This point is where the tangent line gently touches the ellipse.
In simple words: We showed that the given line touches the ellipse using a special rule. Then we found the exact spot where they meet by using a formula for the contact point.

🎯 Exam Tip: When proving a line is tangent, always explicitly state and verify the condition \( c^2 = a^2m^2 + b^2 \). For the point of contact, ensure you use the correct signs for the \( x_1 \) co-ordinate.

Question 4. Find the equation of the tangent to the parabola \( y^2 = 16x \) which is perpendicular to \( 2x + 2y + 3 = 0 \).
Answer: The equation of the parabola is \( y^2 = 16x \).
Comparing this with the standard form \( y^2 = 4ax \), we get \( 4a = 16 \), which means \( a = 4 \).

The given line is \( 2x + 2y + 3 = 0 \).
To find its slope, rearrange it to \( y = mx + c \):
\( 2y = -2x - 3 \)
\( y = -x - \frac{3}{2} \)
The slope of this line is \( m_1 = -1 \).

The required tangent is perpendicular to this line. If two lines are perpendicular, the product of their slopes is -1. So, if \( m_2 \) is the slope of the tangent:
\( m_1 \times m_2 = -1 \)
\( -1 \times m_2 = -1 \)
\( m_2 = 1 \).
So, the slope of the required tangent is \( m = 1 \).

The condition for a line \( y = mx + c \) to be a tangent to the parabola \( y^2 = 4ax \) is \( c = \frac{a}{m} \).
Substitute \( a = 4 \) and \( m = 1 \):
\( c = \frac{4}{1} \)
\( c = 4 \).

Now, we have the slope \( m = 1 \) and the y-intercept \( c = 4 \).
The equation of the tangent line is \( y = mx + c \):
\( y = 1x + 4 \)
\( y = x + 4 \)
Rearranging into the general form:
\( x - y + 4 = 0 \)

This is the equation of the tangent to the parabola that is perpendicular to the given line. It means this line touches the parabola at one point and forms a right angle with the original line.
In simple words: We found the slope of the given line. Since the tangent line is perpendicular, we knew its slope. Then, we used a special formula for parabola tangents to find the full equation of that tangent line.

🎯 Exam Tip: Remember that for perpendicular lines, the product of their slopes is -1. Also, the condition for tangency to a parabola \( y^2 = 4ax \) is \( c = a/m \), which is specific to parabolas and should not be confused with ellipse or hyperbola conditions.

Question 5. Find the equation of the tangent at \( t = 2 \) to the parabola \( y^2 = 8x \) (Hint: use parametric form).
Answer: The equation of the parabola is \( y^2 = 8x \).
Comparing this with the standard form \( y^2 = 4ax \), we get \( 4a = 8 \), which means \( a = 2 \).

We are asked to find the tangent at \( t = 2 \) using the parametric form.
The parametric form for a point on the parabola \( y^2 = 4ax \) is \( (x, y) = (at^2, 2at) \).
Given \( a = 2 \) and \( t = 2 \), let's find the co-ordinates of the point of tangency:
\( x = at^2 = 2(2)^2 = 2(4) = 8 \)
\( y = 2at = 2(2)(2) = 8 \)
So, the point of tangency \( (x_1, y_1) \) is \( (8, 8) \).

The equation of the tangent to the parabola \( y^2 = 4ax \) at a point \( (x_1, y_1) \) is \( yy_1 = 2a(x + x_1) \).
Substitute \( x_1 = 8 \), \( y_1 = 8 \), and \( a = 2 \):
\( y(8) = 2(2)(x + 8) \)
\( 8y = 4(x + 8) \)
Divide both sides by 4 to simplify:
\( \implies 2y = x + 8 \)
Rearrange into the general form:
\( x - 2y + 8 = 0 \)

Alternatively, using the parametric form of the tangent equation:
The equation of the tangent to the parabola \( y^2 = 4ax \) at parameter 't' is \( yt = x + at^2 \).
Substitute \( a = 2 \) and \( t = 2 \):
\( y(2) = x + 2(2)^2 \)
\( 2y = x + 2(4) \)
\( 2y = x + 8 \)
Rearranging into the general form:
\( x - 2y + 8 = 0 \)

Both methods yield the same result. The parametric form is very efficient for finding tangents at specific parameter values. This equation represents the line that touches the parabola exactly at the point corresponding to \( t=2 \).
In simple words: We used a special way to describe points on a parabola using 't'. For \( t=2 \), we found the exact spot on the parabola. Then we used a formula that finds the straight line touching the parabola at that spot.

🎯 Exam Tip: When using parametric form, remember the point is \( (at^2, 2at) \) and the tangent equation is \( yt = x + at^2 \). This approach is often quicker than using \( y_1 = 2a(x + x_1) \) if 't' is given directly.

Question 6. Find the equations of the tangent and normal to hyperbola \( 12x^2 – 9y^2 = 108 \) at \( \theta = \frac{\pi}{3} \). (Hint: use parametric form)
Answer: The given equation of the hyperbola is \( 12x^2 - 9y^2 = 108 \).
To convert it to standard form, divide by 108:
\( \frac{12x^2}{108} - \frac{9y^2}{108} = \frac{108}{108} \)
\( \frac{x^2}{9} - \frac{y^2}{12} = 1 \)
Comparing this with the standard hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we get \( a^2 = 9 \) and \( b^2 = 12 \).
From this, \( a = \sqrt{9} = 3 \) and \( b = \sqrt{12} = 2\sqrt{3} \).
We are given the parameter \( \theta = \frac{\pi}{3} \).

(i) Equation of the tangent to the hyperbola at \( \theta \):
The parametric equation of the tangent to a hyperbola at a point with parameter \( \theta \) is \( \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \).
Substitute \( a = 3 \), \( b = 2\sqrt{3} \), and \( \theta = \frac{\pi}{3} \):
\( \sec \frac{\pi}{3} = 2 \) and \( \tan \frac{\pi}{3} = \sqrt{3} \).
\( \frac{x (2)}{3} - \frac{y (\sqrt{3})}{2\sqrt{3}} = 1 \)
\( \frac{2x}{3} - \frac{y}{2} = 1 \)
To remove the fractions, find the least common multiple of 3 and 2, which is 6. Multiply the entire equation by 6:
\( 6 \left( \frac{2x}{3} \right) - 6 \left( \frac{y}{2} \right) = 6(1) \)
\( 4x - 3y = 6 \)
Rearrange into the general form:
\( 4x - 3y - 6 = 0 \)
This is the equation of the tangent line. It touches the hyperbola at the specific point defined by \( \theta = \frac{\pi}{3} \).

(ii) Equation of the normal to the hyperbola at \( \theta \):
The parametric equation of the normal to a hyperbola at a point with parameter \( \theta \) is \( \frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2 \).
Substitute \( a = 3 \), \( b = 2\sqrt{3} \), \( a^2 = 9 \), \( b^2 = 12 \), and \( \theta = \frac{\pi}{3} \):
\( \frac{3x}{2} + \frac{2\sqrt{3}y}{\sqrt{3}} = 9 + 12 \)
\( \frac{3x}{2} + 2y = 21 \)
To remove the fraction, multiply the entire equation by 2:
\( 2 \left( \frac{3x}{2} \right) + 2(2y) = 2(21) \)
\( 3x + 4y = 42 \)
Rearrange into the general form:
\( 3x + 4y - 42 = 0 \)
This is the equation of the normal line. It passes through the point of tangency and is perpendicular to the tangent line at that point.
In simple words: First, we wrote the hyperbola in its standard form. Then, using special formulas for tangent and normal lines (which depend on a value called theta), we put in the given theta value and the numbers for the hyperbola to get the equations of both lines.

🎯 Exam Tip: Remember to first convert the hyperbola equation to standard form to correctly identify \( a^2 \) and \( b^2 \). Be careful with trigonometric values for \( \sec \theta \) and \( \tan \theta \) at common angles like \( \frac{\pi}{3} \).

Question 7. Prove that the point of intersection of the tangents at 't₁' and 't₂' on the parabola \( y^2 = 4ax \) is \( [at_1t_2, a(t_1 + t_2)] \).
Answer: The equation of the parabola is \( y^2 = 4ax \).
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point with parameter 't' is given by \( yt = x + at^2 \).

For the tangent at 't₁', the equation is:
\( yt_1 = x + at_1^2 \) ... (1)

For the tangent at 't₂', the equation is:
\( yt_2 = x + at_2^2 \) ... (2)

To find the point of intersection, we subtract equation (2) from equation (1):
\( yt_1 - yt_2 = (x + at_1^2) - (x + at_2^2) \)
\( y(t_1 - t_2) = at_1^2 - at_2^2 \)
\( y(t_1 - t_2) = a(t_1^2 - t_2^2) \)
We know that \( t_1^2 - t_2^2 \) can be factored as \( (t_1 + t_2)(t_1 - t_2) \).
\( y(t_1 - t_2) = a(t_1 + t_2)(t_1 - t_2) \)
Assuming \( t_1 \neq t_2 \) (otherwise, it's the same tangent, not an intersection of two distinct tangents), we can divide both sides by \( (t_1 - t_2) \):
\( y = a(t_1 + t_2) \)

Now substitute this value of 'y' back into equation (1) to find 'x':
\( (a(t_1 + t_2))t_1 = x + at_1^2 \)
\( at_1^2 + at_1t_2 = x + at_1^2 \)
Subtract \( at_1^2 \) from both sides:
\( x = at_1t_2 \)

Thus, the point of intersection of the tangents at 't₁' and 't₂' is \( (at_1t_2, a(t_1 + t_2)) \). This shows how the coordinates of the intersection point are directly related to the parameters \( t_1 \) and \( t_2 \) of the tangent points.
In simple words: We wrote down the equations for two tangent lines on a parabola using their special 't' values. By solving these two equations together (like finding where two lines cross), we proved that their meeting point is at \( (at_1t_2, a(t_1 + t_2)) \).

🎯 Exam Tip: Always state the general parametric tangent equation clearly. When solving simultaneous equations, ensure clear steps for isolating 'x' and 'y' and remember the difference of squares factorization for \( t_1^2 - t_2^2 \).

Question 8. If the normal at the point 't₁' on the parabola \( y^2 = 4ax \) meets the parabola again at 't₂', prove that \( t_2 = -(t_1 + \frac{2}{t_1}) \).
Answer: The equation of the parabola is \( y^2 = 4ax \).
The equation of the normal to the parabola \( y^2 = 4ax \) at a point with parameter 't₁' is given by \( y + xt_1 = 2at_1 + at_1^3 \).

This normal line meets the parabola again at a point with parameter 't₂'. The co-ordinates of this point on the parabola are \( (at_2^2, 2at_2) \).
Since this point lies on the normal line, we can substitute its co-ordinates into the normal's equation:
\( (2at_2) + (at_2^2)t_1 = 2at_1 + at_1^3 \)
Divide the entire equation by 'a' (assuming \( a \neq 0 \), which is true for a parabola):
\( 2t_2 + t_1t_2^2 = 2t_1 + t_1^3 \)
Rearrange the terms to group \( t_2 \) terms on one side and \( t_1 \) terms on the other:
\( t_1t_2^2 + 2t_2 = t_1^3 + 2t_1 \)
Factor out \( t_2 \) from the left side and \( t_1 \) from the right side:
\( t_2(t_1t_2 + 2) = t_1(t_1^2 + 2) \)
This step usually leads to a cubic equation. A more straightforward approach is to group the terms differently:
\( 2t_2 - 2t_1 = t_1^3 - t_1t_2^2 \)
Factor out common terms:
\( 2(t_2 - t_1) = t_1(t_1^2 - t_2^2) \)
Use the difference of squares factorization \( t_1^2 - t_2^2 = (t_1 - t_2)(t_1 + t_2) \):
\( 2(t_2 - t_1) = t_1(t_1 - t_2)(t_1 + t_2) \)
We can rewrite \( (t_2 - t_1) \) as \( -(t_1 - t_2) \):
\( -2(t_1 - t_2) = t_1(t_1 - t_2)(t_1 + t_2) \)
If \( t_1 \neq t_2 \) (meaning the normal meets the parabola at a different point), we can divide both sides by \( (t_1 - t_2) \):
\( -2 = t_1(t_1 + t_2) \)
Now, we need to solve for \( t_2 \):
\( -\frac{2}{t_1} = t_1 + t_2 \)
\( t_2 = -\frac{2}{t_1} - t_1 \)
\( t_2 = -(t_1 + \frac{2}{t_1}) \)
This proves the required relationship. This formula shows how the parameter of the second intersection point is determined by the parameter of the first point where the normal is drawn.
In simple words: We took the equation of the normal line at a point \( t_1 \) on the parabola. Then, we made that line intersect the parabola again at point \( t_2 \). By solving the equations, we found a direct link between \( t_1 \) and \( t_2 \), proving that \( t_2 \) can be found from \( t_1 \) using the given formula.

🎯 Exam Tip: When dealing with normals intersecting a curve again, always use the parametric form of both the normal's equation and the point on the curve. Careful factorization of the difference of squares term is key to simplifying the expression.