OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Exercise 25 (B)

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Ex 25(b)

Question 1. Find the tangent to the parabola \( y^2 = 16x \), making an angle of 45° with the x-axis.
Answer: The standard equation for a tangent to a parabola is \( y = mx + \frac{a}{m} \). Here, the slope \( m \) is given by the angle the tangent makes with the x-axis, so \( m = \tan 45° = 1 \).
To find \( a \), we compare the given parabola equation \( y^2 = 16x \) with the general form \( y^2 = 4ax \).
\( \implies 4a = 16 \)
\( \implies a = 4 \)
Now, we substitute the values of \( m = 1 \) and \( a = 4 \) into the tangent equation:
\( y = (1)x + \frac{4}{1} \)
\( \implies y = x + 4 \)
This is the required equation of the tangent to the given parabola. This tangent intersects the parabola at exactly one point.
In simple words: First, find the slope of the tangent using the angle. Then, compare the parabola's equation to the standard form to find the value of 'a'. Finally, put these values into the tangent formula to get the equation.

🎯 Exam Tip: Always remember the standard equation of a tangent to a parabola \( y^2 = 4ax \) is \( y = mx + \frac{a}{m} \). The slope \( m \) is \( \tan \theta \) where \( \theta \) is the angle the tangent makes with the x-axis.

Question 2. A tangent to the parabola \( y^2 = 16x \) makes an angle of 60° with the x-axis. Find its point of contact.
Answer: We know that a line \( y = mx + c \) touches the parabola \( y^2 = 4ax \) at the point \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \).
From the given parabola \( y^2 = 16x \), comparing it with \( y^2 = 4ax \):
\( 4a = 16 \)
\( \implies a = 4 \)
The tangent makes an angle of 60° with the x-axis, so the slope \( m = \tan 60° = \sqrt{3} \).
Now, we substitute \( a = 4 \) and \( m = \sqrt{3} \) into the point of contact formula:
Point of contact \( = \left(\frac{4}{(\sqrt{3})^2}, \frac{2 \times 4}{\sqrt{3}}\right) \)
\( = \left(\frac{4}{3}, \frac{8}{\sqrt{3}}\right) \)
This is the exact point where the tangent line touches the parabola.
In simple words: First, find the 'a' value from the parabola equation and the slope 'm' from the angle. Then, use a special formula that gives you the exact point where the line touches the parabola.

🎯 Exam Tip: Memorize the formula for the point of contact \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \) for a tangent with slope \( m \) to the parabola \( y^2 = 4ax \). This saves time and avoids errors in calculations.

Question 3.
(i) Find the equations of the tangents to the parabola \( y^2 = 6x \) which pass through the point \( \left(\frac{3}{2}, 5\right) \).
(ii) Find the equations of the tangents to the parabola \( y^2 + 12x = 0 \) from the point \( (3, 8) \).
Answer:
(i) For the parabola \( y^2 = 6x \), comparing with \( y^2 = 4ax \):
\( 4a = 6 \)
\( \implies a = \frac{6}{4} = \frac{3}{2} \)
The equation of any tangent to the parabola \( y^2 = 4ax \) is \( y = mx + \frac{a}{m} \).
Substituting \( a = \frac{3}{2} \):
\( y = mx + \frac{3/2}{m} \)
\( \implies y = mx + \frac{3}{2m} \) ...(1)
Since this tangent passes through the point \( \left(\frac{3}{2}, 5\right) \), we substitute \( x = \frac{3}{2} \) and \( y = 5 \) into equation (1):
\( 5 = m\left(\frac{3}{2}\right) + \frac{3}{2m} \)
Multiply the entire equation by \( 2m \) to clear denominators:
\( 10m = 3m^2 + 3 \)
Rearrange into a quadratic equation:
\( 3m^2 - 10m + 3 = 0 \)
Factor the quadratic equation:
\( 3m^2 - 9m - m + 3 = 0 \)
\( 3m(m - 3) - 1(m - 3) = 0 \)
\( (m - 3)(3m - 1) = 0 \)
This gives two possible values for the slope \( m \):
\( m = 3 \) or \( m = \frac{1}{3} \)
Now, substitute these values of \( m \) back into equation (1) to find the tangent equations:
For \( m = 3 \):
\( y = 3x + \frac{3}{2(3)} \)
\( \implies y = 3x + \frac{1}{2} \)
\( \implies 2y = 6x + 1 \) (This is the first tangent equation)
For \( m = \frac{1}{3} \):
\( y = \frac{1}{3}x + \frac{3}{2(\frac{1}{3})} \)
\( \implies y = \frac{1}{3}x + \frac{9}{2} \)
\( \implies 6y = 2x + 27 \) (This is the second tangent equation)
(ii) For the parabola \( y^2 + 12x = 0 \), we rewrite it as \( y^2 = -12x \).
Comparing with \( y^2 = 4ax \):
\( 4a = -12 \)
\( \implies a = -3 \)
The equation of any tangent to the parabola \( y^2 = 4ax \) is \( y = mx + \frac{a}{m} \).
Substituting \( a = -3 \):
\( y = mx + \frac{-3}{m} \)
\( \implies y = mx - \frac{3}{m} \) ...(2)
Since this tangent passes through the point \( (3, 8) \), we substitute \( x = 3 \) and \( y = 8 \) into equation (2):
\( 8 = m(3) - \frac{3}{m} \)
Multiply the entire equation by \( m \) to clear denominators:
\( 8m = 3m^2 - 3 \)
Rearrange into a quadratic equation:
\( 3m^2 - 8m - 3 = 0 \)
Factor the quadratic equation:
\( 3m^2 - 9m + m - 3 = 0 \)
\( 3m(m - 3) + 1(m - 3) = 0 \)
\( (m - 3)(3m + 1) = 0 \)
This gives two possible values for the slope \( m \):
\( m = 3 \) or \( m = -\frac{1}{3} \)
Now, substitute these values of \( m \) back into equation (2) to find the tangent equations:
For \( m = 3 \):
\( y = 3x - \frac{3}{3} \)
\( \implies y = 3x - 1 \) (This is the first tangent equation)
For \( m = -\frac{1}{3} \):
\( y = -\frac{1}{3}x - \frac{3}{-\frac{1}{3}} \)
\( \implies y = -\frac{1}{3}x + 9 \)
\( \implies 3y = -x + 27 \) (This is the second tangent equation)
These equations represent the two lines that touch the parabola and pass through the given point.
In simple words: First, find the value of 'a' from the parabola's equation. Then, write the general tangent equation with 'm' (slope). Since the tangent must pass through a specific point, put the point's coordinates into the tangent equation. This will give you a quadratic equation for 'm'. Solve for 'm' to get two possible slopes. Finally, use each slope in the tangent equation to get the two tangent lines.

🎯 Exam Tip: When finding tangents from an external point, you will typically get a quadratic equation for \( m \), leading to two tangent lines. Always check both values of \( m \) by substituting them back into the general tangent equation.

Question 4. Show that the line \( 12y – 20x – 9 = 0 \) touches the parabola \( y^2 = 5x \).
Answer: We need to show that the given line satisfies the condition for tangency to the parabola.
Given equation of the parabola: \( y^2 = 5x \) ...(1)
Comparing this with the standard form \( y^2 = 4ax \):
\( 4a = 5 \)
\( \implies a = \frac{5}{4} \)
Given equation of the line: \( 12y – 20x – 9 = 0 \)
Rearrange the line equation into the slope-intercept form \( y = mx + c \):
\( 12y = 20x + 9 \)
\( y = \frac{20}{12}x + \frac{9}{12} \)
\( y = \frac{5}{3}x + \frac{3}{4} \) ...(2)
From equation (2), we identify the slope \( m = \frac{5}{3} \) and the y-intercept \( c = \frac{3}{4} \).
For a line \( y = mx + c \) to touch the parabola \( y^2 = 4ax \), the condition is \( c = \frac{a}{m} \).
Let's check if this condition holds:
Calculate \( \frac{a}{m} \):
\( \frac{a}{m} = \frac{5/4}{5/3} \)
\( = \frac{5}{4} \times \frac{3}{5} \)
\( = \frac{3}{4} \)
Since \( c = \frac{3}{4} \) and \( \frac{a}{m} = \frac{3}{4} \), the condition \( c = \frac{a}{m} \) is satisfied. Therefore, the line \( 12y – 20x – 9 = 0 \) touches the parabola \( y^2 = 5x \). This relationship ensures they meet at only one point.
In simple words: First, find 'a' from the parabola equation and 'm' and 'c' from the line equation. Then, check if 'c' is equal to 'a/m'. If they are equal, the line touches the parabola.

🎯 Exam Tip: The tangency condition \( c = \frac{a}{m} \) is crucial for parabolas. Always express the line in \( y = mx + c \) form before comparing to ensure correct identification of \( m \) and \( c \).

Question 5. Show that the line \( x + y = 1 \) touches the parabola \( y = x - x^2 \).
Answer: To show that the line touches the parabola, we substitute the line equation into the parabola equation and check if the resulting quadratic equation has equal roots.
Given equation of the line: \( x + y = 1 \)
We can rewrite this as: \( y = 1 - x \) ...(1)
Given equation of the parabola: \( y = x - x^2 \) ...(2)
Substitute equation (1) into equation (2):
\( 1 - x = x - x^2 \)
Rearrange the terms to form a quadratic equation:
\( x^2 - x - x + 1 = 0 \)
\( x^2 - 2x + 1 = 0 \)
This quadratic equation is a perfect square:
\( (x - 1)^2 = 0 \) ...(3)
Equation (3) has equal roots, \( x = 1 \). When a quadratic equation resulting from the intersection of a line and a curve has equal roots, it means the line is tangent to the curve, touching it at exactly one point. In this case, the line \( x + y = 1 \) touches the parabola \( y = x - x^2 \) at \( x = 1 \).
In simple words: Take the line's equation and put it into the parabola's equation. If the new equation you get has two identical answers for 'x', it means the line just touches the parabola at one point.

🎯 Exam Tip: A key indicator that a line is tangent to a curve is when the discriminant of the resulting quadratic equation (from substitution) is zero, or when it simplifies to a perfect square like \( (x-k)^2 = 0 \).

Question 6. Show that the line \( x + ny + an^2 = 0 \) touches the parabola \( y^2 = 4ax \) and find the point of contact.
Answer: To show tangency, we substitute the line equation into the parabola equation and verify that the resulting quadratic equation has equal roots.
Given equation of the line: \( x + ny + an^2 = 0 \) ...(1)
We can express \( y \) from this equation:
\( ny = -x - an^2 \)
\( y = \frac{-x - an^2}{n} \)
Given equation of the parabola: \( y^2 = 4ax \) ...(2)
Substitute the expression for \( y \) from equation (1) into equation (2):
\( \left(\frac{-x - an^2}{n}\right)^2 = 4ax \)
\( \frac{(x + an^2)^2}{n^2} = 4ax \)
\( (x + an^2)^2 = 4axn^2 \)
Expand the left side:
\( x^2 + 2x(an^2) + (an^2)^2 = 4axn^2 \)
\( x^2 + 2axn^2 + a^2n^4 = 4axn^2 \)
Rearrange into a quadratic equation in \( x \):
\( x^2 + 2axn^2 - 4axn^2 + a^2n^4 = 0 \)
\( x^2 - 2axn^2 + a^2n^4 = 0 \)
This is a perfect square:
\( (x - an^2)^2 = 0 \) ...(3)
Since equation (3) is a quadratic in \( x \) with equal roots, it confirms that the line \( x + ny + an^2 = 0 \) touches the parabola \( y^2 = 4ax \).
To find the point of contact, we use the root of equation (3):
\( x - an^2 = 0 \)
\( \implies x = an^2 \)
Now, substitute this value of \( x \) back into the line equation \( x + ny + an^2 = 0 \):
\( an^2 + ny + an^2 = 0 \)
\( ny + 2an^2 = 0 \)
\( ny = -2an^2 \)
\( y = -2an \)
Thus, the required point of contact is \( (an^2, -2an) \). This point is unique, showing the single intersection.
In simple words: First, rewrite the line equation to express 'y'. Put this 'y' into the parabola equation. If the new equation turns into a perfect square, the line touches the parabola. The value of 'x' from that perfect square gives you one part of the contact point. Use this 'x' back in the line equation to find 'y' for the full contact point.

🎯 Exam Tip: Remember that for a line to touch a parabola, the quadratic equation formed by their intersection must have equal roots. The point of contact is found by solving for \( x \) (or \( y \)) from this quadratic and then finding the corresponding \( y \) (or \( x \)) using the line equation.

Question 7. Find the tangents to the ellipse \( x^2 + 9y^2 = 3 \), which are (i) parallel (ii) perpendicular to the line \( 3x + 4y = 9 \).
Answer: First, let's normalize the ellipse equation and find the properties of the given line.
Given ellipse equation: \( x^2 + 9y^2 = 3 \)
Divide by 3 to get the standard form:
\( \frac{x^2}{3} + \frac{9y^2}{3} = 1 \)
\( \implies \frac{x^2}{3} + \frac{y^2}{1/3} = 1 \) ...(1)
Comparing this with the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):
\( a^2 = 3 \) and \( b^2 = \frac{1}{3} \)
Given line equation: \( 3x + 4y = 9 \) ...(2)
Rewrite in slope-intercept form \( y = mx + c \):
\( 4y = -3x + 9 \)
\( y = -\frac{3}{4}x + \frac{9}{4} \)
The slope of this line is \( m_L = -\frac{3}{4} \).
The general equation of a tangent to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with slope \( m \) is \( y = mx \pm \sqrt{a^2 m^2 + b^2} \).
(i) Tangents parallel to the line \( 3x + 4y = 9 \):
If the tangents are parallel to the given line, their slope \( m \) will be the same as the line's slope.
\( m = m_L = -\frac{3}{4} \)
Substitute \( a^2 = 3 \), \( b^2 = \frac{1}{3} \), and \( m = -\frac{3}{4} \) into the tangent equation:
\( y = \left(-\frac{3}{4}\right)x \pm \sqrt{3\left(-\frac{3}{4}\right)^2 + \frac{1}{3}} \)
\( y = -\frac{3}{4}x \pm \sqrt{3\left(\frac{9}{16}\right) + \frac{1}{3}} \)
\( y = -\frac{3}{4}x \pm \sqrt{\frac{27}{16} + \frac{1}{3}} \)
\( y = -\frac{3}{4}x \pm \sqrt{\frac{81 + 16}{48}} \)
\( y = -\frac{3}{4}x \pm \sqrt{\frac{97}{48}} \)
\( y = -\frac{3}{4}x \pm \frac{\sqrt{97}}{\sqrt{48}} \)
\( y = -\frac{3}{4}x \pm \frac{\sqrt{97}}{4\sqrt{3}} \)
Multiply by \( 4\sqrt{3} \) to clear denominators:
\( 4\sqrt{3}y = -3\sqrt{3}x \pm \sqrt{97} \)
\( 3\sqrt{3}x + 4\sqrt{3}y = \pm \sqrt{97} \) (These are the equations of the tangents parallel to the given line)
(ii) Tangents perpendicular to the line \( 3x + 4y = 9 \):
If the tangents are perpendicular to the given line, their slope \( m \) will be the negative reciprocal of the line's slope.
\( m = -\frac{1}{m_L} = -\frac{1}{(-3/4)} = \frac{4}{3} \)
Substitute \( a^2 = 3 \), \( b^2 = \frac{1}{3} \), and \( m = \frac{4}{3} \) into the tangent equation:
\( y = \left(\frac{4}{3}\right)x \pm \sqrt{3\left(\frac{4}{3}\right)^2 + \frac{1}{3}} \)
\( y = \frac{4}{3}x \pm \sqrt{3\left(\frac{16}{9}\right) + \frac{1}{3}} \)
\( y = \frac{4}{3}x \pm \sqrt{\frac{16}{3} + \frac{1}{3}} \)
\( y = \frac{4}{3}x \pm \sqrt{\frac{17}{3}} \)
Multiply by \( 3 \) to clear denominators:
\( 3y = 4x \pm \sqrt{17 \times 3} \)
\( 3y = 4x \pm \sqrt{51} \) (These are the equations of the tangents perpendicular to the given line)
This demonstrates how the slope affects the tangent equations for an ellipse.
In simple words: First, get the ellipse into its standard form to find 'a²' and 'b²'. Then, find the slope of the given line. For parallel tangents, use the same slope. For perpendicular tangents, use the negative reciprocal slope. Finally, plug 'a²', 'b²', and the correct 'm' into the general tangent formula for an ellipse to get the equations.

🎯 Exam Tip: Remember the condition for parallel lines (slopes are equal) and perpendicular lines (product of slopes is -1). The formula for tangents to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( y = mx \pm \sqrt{a^2 m^2 + b^2} \). Don't confuse it with the hyperbola formula.

Question 8. Find the equations of the tangents to the ellipse \( \frac{x^2}{2} + \frac{y^2}{7} = 1 \) that make an angle of 45° with the x-axis.
Answer: Given equation of the ellipse: \( \frac{x^2}{2} + \frac{y^2}{7} = 1 \) ...(1)
This is a vertical ellipse since the denominator of \( y^2 \) (which is 7) is greater than the denominator of \( x^2 \) (which is 2).
Comparing with the standard form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \), where \( a > b \):
\( b^2 = 2 \) and \( a^2 = 7 \)
The tangents make an angle of 45° with the x-axis. The slope \( m \) is given by:
\( m = \tan 45° = 1 \)
The general equation of a tangent to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with slope \( m \) is \( y = mx \pm \sqrt{a^2 m^2 + b^2} \).
Substitute \( a^2 = 7 \), \( b^2 = 2 \), and \( m = 1 \) into the tangent equation:
\( y = (1)x \pm \sqrt{7(1)^2 + 2} \)
\( y = x \pm \sqrt{7 + 2} \)
\( y = x \pm \sqrt{9} \)
\( y = x \pm 3 \)
Therefore, the two equations of the tangents are \( y = x + 3 \) and \( y = x - 3 \). These lines will touch the ellipse at exactly two points.
In simple words: First, identify 'a²' and 'b²' from the ellipse equation. Then, find the slope 'm' using the given angle. Finally, put these values into the ellipse tangent formula. This will give you two lines that touch the ellipse at that specific angle.

🎯 Exam Tip: Be careful to correctly identify \( a^2 \) and \( b^2 \) when the ellipse is vertical (i.e., \( a^2 \) is under \( y^2 \)). The general tangent formula remains the same, but the values for \( a^2 \) and \( b^2 \) are swapped compared to a horizontal ellipse where \( a^2 \) is under \( x^2 \).

Question 9. Find the equation of the tangents to the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), which make equal intercepts on the axes.
Answer: Given equation of the ellipse: \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) ...(1)
Comparing this with the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):
\( a^2 = 16 \) and \( b^2 = 9 \)
When a line makes equal intercepts on the axes, its equation can be written as \( \frac{x}{k} + \frac{y}{k} = 1 \), which simplifies to \( x + y = k \), or \( y = -x + k \). This means the slope \( m \) of such a line is \( -1 \).
However, tangents can also make intercepts that are equal in magnitude but opposite in sign, like \( \frac{x}{k} + \frac{y}{-k} = 1 \), meaning \( x - y = k \) or \( y = x - k \). In this case, the slope \( m \) is \( 1 \).
So, the slope of tangents making equal intercepts (in magnitude) on the axes is \( m = \pm 1 \).
The general equation of a tangent to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with slope \( m \) is \( y = mx \pm \sqrt{a^2 m^2 + b^2} \).
Substitute \( a^2 = 16 \), \( b^2 = 9 \), and \( m = \pm 1 \) into the tangent equation:
\( y = (\pm 1)x \pm \sqrt{16(\pm 1)^2 + 9} \)
\( y = \pm x \pm \sqrt{16(1) + 9} \)
\( y = \pm x \pm \sqrt{25} \)
\( y = \pm x \pm 5 \)
This results in four possible tangent equations:
1. \( y = x + 5 \)
2. \( y = x - 5 \)
3. \( y = -x + 5 \)
4. \( y = -x - 5 \)
These lines all have slopes of \( 1 \) or \( -1 \), ensuring equal intercepts on the coordinate axes.
In simple words: First, find 'a²' and 'b²' from the ellipse. If a line makes equal intercepts on the axes, its slope 'm' can be either 1 or -1. Then, use this 'm' along with 'a²' and 'b²' in the ellipse tangent formula. This will give you the equations of the lines that touch the ellipse and have equal intercepts.

🎯 Exam Tip: When a line makes "equal intercepts" on the axes, it typically implies \( |x_{intercept}| = |y_{intercept}| \), which means the slope is \( \pm 1 \). Be sure to consider both positive and negative slopes to find all possible tangent lines.

Question 10. Find the value of 'c' so that \( 2x - y + c = 0 \) may touch the ellipse \( x^2 + 2y^2 = 2 \).
Answer: For a line to touch an ellipse, there's a specific condition involving its slope and y-intercept, and the ellipse's parameters.
Given equation of the line: \( 2x - y + c = 0 \)
Rewrite in slope-intercept form \( y = mx + c' \):
\( y = 2x + c \) ...(1)
From this, we have the slope \( m = 2 \) and the y-intercept is \( c \).
Given equation of the ellipse: \( x^2 + 2y^2 = 2 \)
Divide by 2 to get the standard form:
\( \frac{x^2}{2} + \frac{2y^2}{2} = 1 \)
\( \implies \frac{x^2}{2} + \frac{y^2}{1} = 1 \) ...(2)
Comparing this with the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):
\( a^2 = 2 \) and \( b^2 = 1 \)
The condition for a line \( y = mx + c \) to touch the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( c^2 = a^2 m^2 + b^2 \), or \( c = \pm \sqrt{a^2 m^2 + b^2} \).
Substitute the values \( a^2 = 2 \), \( b^2 = 1 \), and \( m = 2 \) into this condition:
\( c = \pm \sqrt{2(2)^2 + 1} \)
\( c = \pm \sqrt{2(4) + 1} \)
\( c = \pm \sqrt{8 + 1} \)
\( c = \pm \sqrt{9} \)
\( c = \pm 3 \)
Thus, the value of \( c \) can be \( 3 \) or \( -3 \) for the line to be tangent to the ellipse. This means there are two such lines.
In simple words: First, find 'a²' and 'b²' from the ellipse's equation, and the slope 'm' from the line. Then, use the special formula \( c = \pm \sqrt{a^2 m^2 + b^2} \) to find the possible values for 'c'.

🎯 Exam Tip: The tangency condition \( c^2 = a^2 m^2 + b^2 \) (or \( c = \pm \sqrt{a^2 m^2 + b^2} \)) is fundamental for ellipses. Ensure the line is in \( y = mx + c \) form and the ellipse in standard form before applying the condition.

Question 11. Show that the line \( lx + my = 1 \) will touch the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) if \( a^2l^2 + b^2m^2 = 1 \).
Answer: To show that the line touches the ellipse, we need to substitute the line equation into the ellipse equation and demonstrate that the resulting quadratic equation in \( x \) (or \( y \)) has equal roots (i.e., its discriminant is zero).
Given equation of the line: \( lx + my = 1 \) ...(1)
From this, we can express \( y \):
\( my = 1 - lx \)
\( y = \frac{1 - lx}{m} \)
Given equation of the ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) ...(2)
Substitute the expression for \( y \) into the ellipse equation:
\( \frac{x^2}{a^2} + \frac{1}{b^2} \left(\frac{1 - lx}{m}\right)^2 = 1 \)
\( \frac{x^2}{a^2} + \frac{(1 - lx)^2}{b^2m^2} = 1 \)
Multiply the entire equation by \( a^2 b^2 m^2 \) to clear denominators:
\( b^2m^2 x^2 + a^2(1 - lx)^2 = a^2b^2m^2 \)
Expand the term \( (1 - lx)^2 \):
\( b^2m^2 x^2 + a^2(1 - 2lx + l^2x^2) = a^2b^2m^2 \)
\( b^2m^2 x^2 + a^2 - 2a^2lx + a^2l^2x^2 = a^2b^2m^2 \)
Rearrange into a quadratic equation in \( x \), in the form \( Ax^2 + Bx + C = 0 \):
\( (b^2m^2 + a^2l^2)x^2 - (2a^2l)x + (a^2 - a^2b^2m^2) = 0 \) ...(3)
For the line to touch the ellipse, this quadratic equation must have equal roots. This means its discriminant \( D \) must be zero (\( B^2 - 4AC = 0 \)).
Here, \( A = (b^2m^2 + a^2l^2) \), \( B = (-2a^2l) \), and \( C = (a^2 - a^2b^2m^2) \).
\( D = (-2a^2l)^2 - 4(b^2m^2 + a^2l^2)(a^2 - a^2b^2m^2) = 0 \)
\( 4a^4l^2 - 4(b^2m^2 + a^2l^2)a^2(1 - b^2m^2) = 0 \)
Divide by \( 4a^2 \) (assuming \( a \neq 0 \)):
\( a^2l^2 - (b^2m^2 + a^2l^2)(1 - b^2m^2) = 0 \)
\( a^2l^2 - [b^2m^2(1 - b^2m^2) + a^2l^2(1 - b^2m^2)] = 0 \)
\( a^2l^2 - [b^2m^2 - b^4m^4 + a^2l^2 - a^2l^2b^2m^2] = 0 \)
\( a^2l^2 - b^2m^2 + b^4m^4 - a^2l^2 + a^2l^2b^2m^2 = 0 \)
Combine like terms:
\( -b^2m^2 + b^4m^4 + a^2l^2b^2m^2 = 0 \)
Factor out \( b^2m^2 \) (assuming \( b \neq 0 \) and \( m \neq 0 \)):
\( b^2m^2(-1 + b^2m^4 + a^2l^2) = 0 \)
Wait, there's a calculation error in my thought process or the source's intermediate steps. Let's restart from \( a^2l^2 - (b^2m^2 + a^2l^2)(1 - b^2m^2) = 0 \).
\( a^2l^2 - (b^2m^2 - b^4m^4 + a^2l^2 - a^2l^2b^2m^2) = 0 \)
\( a^2l^2 - b^2m^2 + b^4m^4 - a^2l^2 + a^2l^2b^2m^2 = 0 \)
This simplifies to:
\( -b^2m^2 + b^4m^4 + a^2l^2b^2m^2 = 0 \)
Factor out \( b^2m^2 \):
\( b^2m^2 (-1 + b^2m^2 + a^2l^2) = 0 \)
Since \( b^2m^2 \neq 0 \):
\( -1 + b^2m^2 + a^2l^2 = 0 \)
\( \implies a^2l^2 + b^2m^2 = 1 \)
This matches the required condition. This condition ensures that the line is tangential, meeting the ellipse at a single point.
In simple words: To prove the line touches the ellipse, first express 'y' from the line equation and substitute it into the ellipse equation. This will give a quadratic equation in 'x'. For the line to be a tangent, this quadratic equation must have only one solution, meaning its discriminant (B²-4AC) must be zero. When you set the discriminant to zero and simplify, you will find the condition \( a^2l^2 + b^2m^2 = 1 \).

🎯 Exam Tip: The condition \( a^2l^2 + b^2m^2 = 1 \) for a line \( lx + my = 1 \) to touch an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is a standard result. Deriving it by setting the discriminant to zero is a common proof technique.

Question 12. Show that the following lines are tangents to the given hyperbola and determine the points of contact.
(i) \( x + 1 = 0, 4x^2 – 3y^2 = 4 \)
(ii) \( x – 2y + 1 = 0, x^2 – 6y^2 = 3 \)
Answer:
(i) Given equation of the line: \( x + 1 = 0 \)
\( \implies x = -1 \) ...(1)
Given equation of the hyperbola: \( 4x^2 – 3y^2 = 4 \)
Divide by 4 to get the standard form:
\( \frac{4x^2}{4} - \frac{3y^2}{4} = 1 \)
\( \implies \frac{x^2}{1} - \frac{y^2}{4/3} = 1 \) ...(2)
To show tangency, substitute equation (1) into equation (2):
\( \frac{(-1)^2}{1} - \frac{y^2}{4/3} = 1 \)
\( 1 - \frac{y^2}{4/3} = 1 \)
\( \implies -\frac{y^2}{4/3} = 0 \)
\( \implies y^2 = 0 \)
\( \implies y = 0 \)
Since \( y = 0 \) is a double root, the line touches the hyperbola. This confirms it's a tangent.
The point of contact is \( (x, y) = (-1, 0) \).
(ii) Given equation of the line: \( x – 2y + 1 = 0 \)
From this, express \( x \):
\( x = 2y - 1 \) ...(1)
Given equation of the hyperbola: \( x^2 – 6y^2 = 3 \) ...(2)
To show tangency, substitute equation (1) into equation (2):
\( (2y - 1)^2 - 6y^2 = 3 \)
Expand \( (2y - 1)^2 \):
\( 4y^2 - 4y + 1 - 6y^2 = 3 \)
Combine like terms:
\( -2y^2 - 4y + 1 = 3 \)
\( -2y^2 - 4y - 2 = 0 \)
Divide by \( -2 \):
\( y^2 + 2y + 1 = 0 \)
This is a perfect square:
\( (y + 1)^2 = 0 \)
Since \( y = -1 \) is a double root, the line touches the hyperbola.
To find the point of contact, substitute \( y = -1 \) back into equation (1):
\( x = 2(-1) - 1 \)
\( x = -2 - 1 \)
\( x = -3 \)
The point of contact is \( (x, y) = (-3, -1) \). Each part of the problem shows a unique point of intersection.
In simple words: To show a line touches a hyperbola, substitute the line's equation into the hyperbola's. If the resulting equation has only one repeating answer (a double root), then it's a tangent. That single answer gives you one coordinate of the contact point. Use it in the line equation to find the other coordinate.

🎯 Exam Tip: When proving tangency, direct substitution and checking for equal roots (or a zero discriminant) is a reliable method. Always remember to also find the point of contact by solving for the coordinates once tangency is established.

Question 13. Find the equations of the tangents to the hyperbola \( 2x^2 – 3y^2 = 6 \), which are parallel to the line \( x + y – 2 = 0 \).
Answer: First, we need to standardize the hyperbola equation and find the slope of the given line.
Given hyperbola equation: \( 2x^2 – 3y^2 = 6 \)
Divide by 6 to get the standard form:
\( \frac{2x^2}{6} - \frac{3y^2}{6} = 1 \)
\( \implies \frac{x^2}{3} - \frac{y^2}{2} = 1 \) ...(1)
Comparing this with the standard hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \):
\( a^2 = 3 \) and \( b^2 = 2 \)
Given line equation: \( x + y – 2 = 0 \)
Rewrite in slope-intercept form \( y = mx + c \):
\( y = -x + 2 \)
The slope of this line is \( m_L = -1 \).
Since the tangents are parallel to this line, their slope \( m \) will be the same as the line's slope:
\( m = m_L = -1 \)
The general equation of a tangent to a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with slope \( m \) is \( y = mx \pm \sqrt{a^2 m^2 - b^2} \).
Substitute \( a^2 = 3 \), \( b^2 = 2 \), and \( m = -1 \) into the tangent equation:
\( y = (-1)x \pm \sqrt{3(-1)^2 - 2} \)
\( y = -x \pm \sqrt{3(1) - 2} \)
\( y = -x \pm \sqrt{3 - 2} \)
\( y = -x \pm \sqrt{1} \)
\( y = -x \pm 1 \)
This gives two equations for the tangents:
1. \( y = -x + 1 \implies x + y - 1 = 0 \)
2. \( y = -x - 1 \implies x + y + 1 = 0 \)
These two lines are parallel to the given line and touch the hyperbola.
In simple words: First, find 'a²' and 'b²' from the hyperbola's equation and the slope of the given line. Since the tangents are parallel, they have the same slope. Then, use this slope, 'a²', and 'b²' in the hyperbola tangent formula to get the two tangent equations.

🎯 Exam Tip: For parallel tangents, the slopes are identical. Ensure you use the correct tangent condition for a hyperbola, which has a minus sign under the square root, i.e., \( y = mx \pm \sqrt{a^2 m^2 - b^2} \).

Question 14. The tangents from P to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are mutually perpendicular, show that the locus of P is the circle \( x^2 + y^2 = a^2 – b^2 \).
Answer: Let the point \( P \) be \( (x_1, y_1) \). The equation of a tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with slope \( m \) is given by:
\( y = mx \pm \sqrt{a^2 m^2 - b^2} \)
Since the tangent passes through \( P(x_1, y_1) \), we substitute these coordinates into the tangent equation:
\( y_1 = mx_1 \pm \sqrt{a^2 m^2 - b^2} \)
Isolate the square root term:
\( y_1 - mx_1 = \pm \sqrt{a^2 m^2 - b^2} \)
Square both sides to eliminate the square root:
\( (y_1 - mx_1)^2 = a^2 m^2 - b^2 \)
Expand the left side:
\( y_1^2 - 2mx_1y_1 + m^2x_1^2 = a^2 m^2 - b^2 \)
Rearrange this into a quadratic equation in \( m \):
\( m^2x_1^2 - a^2 m^2 - 2mx_1y_1 + y_1^2 + b^2 = 0 \)
\( (x_1^2 - a^2)m^2 - (2x_1y_1)m + (y_1^2 + b^2) = 0 \) ...(1)
Let \( m_1 \) and \( m_2 \) be the two roots of this quadratic equation, which represent the slopes of the two tangents from point \( P \).
The problem states that these tangents are mutually perpendicular. For two lines to be perpendicular, the product of their slopes must be \( -1 \):
\( m_1 m_2 = -1 \)
From the quadratic equation (1), the product of the roots is given by \( \frac{C}{A} \), where \( A = (x_1^2 - a^2) \) and \( C = (y_1^2 + b^2) \).
So, \( m_1 m_2 = \frac{y_1^2 + b^2}{x_1^2 - a^2} \)
Set this equal to \( -1 \):
\( \frac{y_1^2 + b^2}{x_1^2 - a^2} = -1 \)
\( y_1^2 + b^2 = -(x_1^2 - a^2) \)
\( y_1^2 + b^2 = -x_1^2 + a^2 \)
Rearrange to find the locus of point \( P \):
\( x_1^2 + y_1^2 = a^2 - b^2 \)
Replacing \( (x_1, y_1) \) with \( (x, y) \), the locus of point \( P \) is the circle \( x^2 + y^2 = a^2 - b^2 \). This circle is known as the Director Circle of the hyperbola, where all perpendicular tangents intersect.
In simple words: Imagine a point P. Two lines (tangents) from P touch the hyperbola, and these lines are at 90 degrees to each other. We use the general formula for a tangent and make it pass through P. This gives us an equation for the slope 'm'. Since the tangents are perpendicular, the product of their slopes must be -1. Using this, we find that the coordinates of P must follow the equation \( x^2 + y^2 = a^2 - b^2 \).

🎯 Exam Tip: The Director Circle is a crucial concept for both ellipses and hyperbolas. For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), its equation is \( x^2 + y^2 = a^2 + b^2 \). For a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), it is \( x^2 + y^2 = a^2 - b^2 \). Be careful with the sign difference.

Question 15. Show that the straight line \( x + y = 1 \) touches the hyperbola \( 2x^2 – 3y^2 = 6 \). Also find the coordinates of the point of contact.
Answer: To show that the line touches the hyperbola, we will substitute the line equation into the hyperbola equation and verify that the resulting quadratic equation has equal roots.
Given equation of the straight line: \( x + y = 1 \) ...(1)
From this, we can express \( y \):
\( y = 1 - x \)
Given equation of the hyperbola: \( 2x^2 – 3y^2 = 6 \) ...(2)
Substitute the expression for \( y \) from the line equation into the hyperbola equation:
\( 2x^2 - 3(1 - x)^2 = 6 \)
Expand the term \( (1 - x)^2 \):
\( 2x^2 - 3(1 - 2x + x^2) = 6 \)
Distribute the \( -3 \):
\( 2x^2 - 3 + 6x - 3x^2 = 6 \)
Combine like terms:
\( -x^2 + 6x - 3 = 6 \)
Move all terms to one side to form a quadratic equation:
\( -x^2 + 6x - 3 - 6 = 0 \)
\( -x^2 + 6x - 9 = 0 \)
Multiply by \( -1 \) to make the leading coefficient positive:
\( x^2 - 6x + 9 = 0 \)
This is a perfect square trinomial:
\( (x - 3)^2 = 0 \)
Since this quadratic equation has equal roots (\( x = 3 \)), it proves that the line \( x + y = 1 \) touches the hyperbola \( 2x^2 – 3y^2 = 6 \).
To find the coordinates of the point of contact, use the value \( x = 3 \) in the line equation \( y = 1 - x \):
\( y = 1 - 3 \)
\( y = -2 \)
Therefore, the coordinates of the point of contact are \( (3, -2) \). This point is the unique intersection of the line and hyperbola.
In simple words: To check if the line touches the hyperbola, substitute the line's 'y' value into the hyperbola's equation. If the final equation has one repeated answer for 'x', it means the line is a tangent. This 'x' value, along with the 'y' value you find by putting 'x' back into the line equation, gives you the exact point where they touch.

🎯 Exam Tip: When showing tangency, look for a quadratic equation with equal roots \( (x-k)^2 = 0 \). The value of \( k \) is the x-coordinate of the point of contact, which you then use in the line equation to find the y-coordinate.

Question 16. Find the equations of the tangents to the hyperbola \( 4x^2 – 9y^2 = 144 \), which are perpendicular to the line \( 6x + 5y = 21 \).
Answer: First, we need to convert the hyperbola equation into its standard form and determine the slope of the given line.
Given hyperbola equation: \( 4x^2 – 9y^2 = 144 \)
Divide by 144 to get the standard form:
\( \frac{4x^2}{144} - \frac{9y^2}{144} = 1 \)
\( \implies \frac{x^2}{36} - \frac{y^2}{16} = 1 \) ...(1)
Comparing this with the standard hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \):
\( a^2 = 36 \) and \( b^2 = 16 \)
Given line equation: \( 6x + 5y = 21 \)
Rewrite in slope-intercept form \( y = mx + c \):
\( 5y = -6x + 21 \)
\( y = -\frac{6}{5}x + \frac{21}{5} \)
The slope of this line is \( m_L = -\frac{6}{5} \).
Since the tangents are perpendicular to this line, their slope \( m \) will be the negative reciprocal of the line's slope:
\( m = -\frac{1}{m_L} = -\frac{1}{(-6/5)} = \frac{5}{6} \)
The general equation of a tangent to a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with slope \( m \) is \( y = mx \pm \sqrt{a^2 m^2 - b^2} \).
Substitute \( a^2 = 36 \), \( b^2 = 16 \), and \( m = \frac{5}{6} \) into the tangent equation:
\( y = \left(\frac{5}{6}\right)x \pm \sqrt{36\left(\frac{5}{6}\right)^2 - 16} \)
\( y = \frac{5}{6}x \pm \sqrt{36\left(\frac{25}{36}\right) - 16} \)
\( y = \frac{5}{6}x \pm \sqrt{25 - 16} \)
\( y = \frac{5}{6}x \pm \sqrt{9} \)
\( y = \frac{5}{6}x \pm 3 \)
To eliminate the fraction, multiply the entire equation by 6:
\( 6y = 5x \pm 18 \)
This gives two equations for the tangents:
1. \( 6y = 5x + 18 \implies 5x - 6y + 18 = 0 \)
2. \( 6y = 5x - 18 \implies 5x - 6y - 18 = 0 \)
These two lines are perpendicular to the given line and touch the hyperbola.
In simple words: First, rewrite the hyperbola equation into its standard form to find 'a²' and 'b²'. Then, get the slope of the given line. For perpendicular tangents, use the negative reciprocal of that slope. Finally, use this new slope, 'a²', and 'b²' in the hyperbola tangent formula to find the two tangent equations.

🎯 Exam Tip: Remember to correctly calculate the perpendicular slope as the negative reciprocal. The tangent condition for a hyperbola uses a minus sign under the square root, \( \sqrt{a^2 m^2 - b^2} \), which is different from an ellipse.