RBSE Solutions Class 11 Maths Chapter 12 Conic Section Exercise 12.6

Question 1. Prove that line \( y = x + \sqrt{\frac{5}{6}} \) touches ellipse \( 2x^2 + 3y^2 = 1 \). Also find the coordinates of tangent point.
Answer:
The equation of the given ellipse is \( 2x^2 + 3y^2 = 1 \).
We can rewrite this as \( \frac{x^2}{1/2} + \frac{y^2}{1/3} = 1 \).
So, we have \( a^2 = \frac{1}{2} \) and \( b^2 = \frac{1}{3} \).
The equation of the given line is \( y = x + \sqrt{\frac{5}{6}} \).
Comparing this with \( y = mx + c \), we find that \( m = 1 \) and \( c = \sqrt{\frac{5}{6}} \).
The condition for a line \( y = mx + c \) to touch an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( c^2 = a^2m^2 + b^2 \).
Let's check if this condition holds true:
LHS: \( c^2 = \left(\sqrt{\frac{5}{6}}\right)^2 = \frac{5}{6} \)
RHS: \( a^2m^2 + b^2 = \left(\frac{1}{2}\right)(1)^2 + \left(\frac{1}{3}\right) = \frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6} \)
Since LHS = RHS, the condition is satisfied. Therefore, the given line touches the ellipse.

To find the coordinates of the tangent point \( (x_1, y_1) \), we use the formulas:
\( x_1 = \frac{-a^2m}{c} \) and \( y_1 = \frac{b^2}{c} \)
Substitute the values:
\( x_1 = \frac{-(1/2)(1)}{\sqrt{5/6}} = \frac{-1/2}{\sqrt{5/6}} = -\frac{1}{2} \times \sqrt{\frac{6}{5}} = -\frac{\sqrt{6}}{2\sqrt{5}} = -\frac{\sqrt{6}\sqrt{5}}{2\sqrt{5}\sqrt{5}} = -\frac{\sqrt{30}}{10} \)
\( y_1 = \frac{1/3}{\sqrt{5/6}} = \frac{1}{3} \times \sqrt{\frac{6}{5}} = \frac{\sqrt{6}}{3\sqrt{5}} = \frac{\sqrt{6}\sqrt{5}}{3\sqrt{5}\sqrt{5}} = \frac{\sqrt{30}}{15} \)
So, the coordinates of the tangent point are \( \left(-\frac{\sqrt{30}}{10}, \frac{\sqrt{30}}{15}\right) \).
In simple words: We first checked if the line touches the ellipse by using a special math rule. It did touch. Then we used another rule to find the exact spot where they meet, giving us the coordinates.

🎯 Exam Tip: Remember the tangency condition \( c^2 = a^2m^2 + b^2 \) for an ellipse. Always convert the ellipse equation to standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) to correctly identify \( a^2 \) and \( b^2 \).

Question 2. Show that line \( x - 3y - 4 = 0 \) touches ellipse \( 3x^2 + 4y^2 = 20 \).
Answer:
The given equation of the line is \( x - 3y - 4 = 0 \).
We can write this as \( x = 3y + 4 \). (Equation 1)
The given equation of the ellipse is \( 3x^2 + 4y^2 = 20 \). (Equation 2)
To check if the line touches the ellipse, we substitute the value of \( x \) from Equation 1 into Equation 2:
\( 3(3y + 4)^2 + 4y^2 = 20 \)
\( 3(9y^2 + 24y + 16) + 4y^2 = 20 \)
\( 27y^2 + 72y + 48 + 4y^2 = 20 \)
\( 31y^2 + 72y + 48 - 20 = 0 \)
\( 31y^2 + 72y + 28 = 0 \)
This is a quadratic equation in the form \( Ay^2 + By + C = 0 \), where \( A = 31 \), \( B = 72 \), and \( C = 28 \).
For a line to be tangent (touch) an ellipse, the discriminant \( D = B^2 - 4AC \) of the resulting quadratic equation must be zero.
Let's calculate the discriminant:
\( D = (72)^2 - 4(31)(28) \)
\( D = 5184 - 3472 \)
\( D = 1712 \)
Since \( D = 1712 \) and \( 1712 \neq 0 \), the line intersects the ellipse at two distinct points, or does not intersect it at all. Therefore, the given line does not touch the ellipse.
In simple words: We put the line's equation into the ellipse's equation and got a new equation. We then checked a special number called the discriminant. If this number was zero, the line would touch the ellipse. Since it wasn't zero, the line does not touch the ellipse.

🎯 Exam Tip: To show if a line touches a conic section (like an ellipse), substitute the linear equation into the conic equation. If the resulting quadratic equation has a discriminant of zero, the line is tangent. If the discriminant is not zero, it's not tangent.

Question 3. For which value of k, the line \( 3x - 4y = k \) touches the ellipse \( 5x^2 + 4y^2 = 20 \).
Answer:
The equation of the given line is \( 3x - 4y = k \).
We can rewrite this to express \( y \) in terms of \( x \):
\( 4y = 3x - k \)
\( \implies y = \frac{3}{4}x - \frac{k}{4} \)
The equation of the given ellipse is \( 5x^2 + 4y^2 = 20 \).
Substitute the expression for \( y \) from the line equation into the ellipse equation:
\( 5x^2 + 4\left(\frac{3x - k}{4}\right)^2 = 20 \)
\( 5x^2 + 4\frac{(3x - k)^2}{16} = 20 \)
\( 5x^2 + \frac{9x^2 - 6kx + k^2}{4} = 20 \)
Multiply the entire equation by 4 to remove the fraction:
\( 20x^2 + 9x^2 - 6kx + k^2 = 80 \)
\( 29x^2 - 6kx + (k^2 - 80) = 0 \)
For the line to touch the ellipse, the discriminant of this quadratic equation must be zero.
The quadratic equation is in the form \( Ax^2 + Bx + C = 0 \), where \( A = 29 \), \( B = -6k \), and \( C = k^2 - 80 \).
Set the discriminant \( D = B^2 - 4AC \) to zero:
\( (-6k)^2 - 4(29)(k^2 - 80) = 0 \)
\( 36k^2 - 116(k^2 - 80) = 0 \)
\( 36k^2 - 116k^2 + 116 \times 80 = 0 \)
\( -80k^2 + 9280 = 0 \)
\( 80k^2 = 9280 \)
\( k^2 = \frac{9280}{80} \)
\( k^2 = 116 \)
\( k = \pm \sqrt{116} \)
\( k = \pm \sqrt{4 \times 29} \)
\( k = \pm 2\sqrt{29} \)
So, the line touches the ellipse when \( k = \pm 2\sqrt{29} \).
In simple words: We put the line's equation into the ellipse's equation, which gave us a quadratic equation with 'k' in it. For the line to just touch, a special value called the discriminant must be zero. We set the discriminant to zero and solved for 'k' to find the required values.

🎯 Exam Tip: When finding a constant for tangency, using the discriminant \( B^2 - 4AC = 0 \) is a reliable method after substituting the line equation into the conic section equation. Alternatively, you could use the direct tangency condition formula \( c^2 = a^2m^2 + b^2 \) for the ellipse after converting both equations to standard forms.

Question 5. Find the condition that line \( lx + my = n \) touches the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Answer:
The equation of the given line is \( lx + my = n \).
We can rewrite this in the slope-intercept form \( y = M x + C \):
\( my = -lx + n \)
\( \implies y = -\frac{l}{m}x + \frac{n}{m} \)
Comparing this with \( y = M x + C \), we get the slope \( M = -\frac{l}{m} \) and the y-intercept \( C = \frac{n}{m} \).
The equation of the given ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
The condition for a line \( y = M x + C \) to touch an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by \( C^2 = a^2M^2 + b^2 \).
Substitute the values of \( M \) and \( C \) into this condition:
\( \left(\frac{n}{m}\right)^2 = a^2\left(-\frac{l}{m}\right)^2 + b^2 \)
\( \implies \frac{n^2}{m^2} = a^2\frac{l^2}{m^2} + b^2 \)
Now, multiply the entire equation by \( m^2 \) to clear the denominators:
\( n^2 = a^2l^2 + b^2m^2 \)
This is the required condition for the line \( lx + my = n \) to touch the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
In simple words: We rewrote the line's equation to find its slope and where it crosses the y-axis. Then, we used a standard formula that tells us when a line touches an ellipse, plugging in the values from both the line and the ellipse. This gave us a simple rule, \( n^2 = a^2l^2 + b^2m^2 \), that must be true for the line to touch the ellipse.

🎯 Exam Tip: Memorize the standard tangency condition \( c^2 = a^2m^2 + b^2 \) for an ellipse. For general line equations, convert them to slope-intercept form \( y = mx + c \) to directly apply the formula.

Question 6. Find the equations of tangent for ellipse \( 4x^2 + 3y^2 = 5 \) which makes equation angle of \( 60^\circ \) with x-axis. Also find the coordinates of tangent point.
Answer:
The equation of the given ellipse is \( 4x^2 + 3y^2 = 5 \).
To convert it to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), divide by 5:
\( \frac{4x^2}{5} + \frac{3y^2}{5} = 1 \)
\( \implies \frac{x^2}{5/4} + \frac{y^2}{5/3} = 1 \)
So, we have \( a^2 = \frac{5}{4} \) and \( b^2 = \frac{5}{3} \).
The tangent line makes an angle of \( 60^\circ \) with the x-axis.
The slope of the tangent line is \( m = \tan 60^\circ = \sqrt{3} \).
The equation of a tangent to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with slope \( m \) is given by the formula:
\( y = mx \pm \sqrt{a^2m^2 + b^2} \)
Substitute the values of \( a^2, b^2 \), and \( m \):
\( y = \sqrt{3}x \pm \sqrt{\left(\frac{5}{4}\right)(\sqrt{3})^2 + \left(\frac{5}{3}\right)} \)
\( y = \sqrt{3}x \pm \sqrt{\left(\frac{5}{4}\right)(3) + \frac{5}{3}} \)
\( y = \sqrt{3}x \pm \sqrt{\frac{15}{4} + \frac{5}{3}} \)
To add the fractions, find a common denominator, which is 12:
\( y = \sqrt{3}x \pm \sqrt{\frac{15 \times 3}{12} + \frac{5 \times 4}{12}} \)
\( y = \sqrt{3}x \pm \sqrt{\frac{45 + 20}{12}} \)
\( y = \sqrt{3}x \pm \sqrt{\frac{65}{12}} \)
We can simplify \( \sqrt{12} \) as \( \sqrt{4 \times 3} = 2\sqrt{3} \):
\( y = \sqrt{3}x \pm \frac{\sqrt{65}}{2\sqrt{3}} \)
These are the two equations of the tangent lines.

Now, let's find the coordinates of the tangent points \( (x_1, y_1) \).
For a tangent \( y = mx + c \) to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the coordinates of the point of tangency are:
\( x_1 = \frac{-a^2m}{c} \) and \( y_1 = \frac{b^2}{c} \)
Here, \( m = \sqrt{3} \), \( a^2 = \frac{5}{4} \), \( b^2 = \frac{5}{3} \), and \( c = \pm \frac{\sqrt{65}}{2\sqrt{3}} \).

Case 1: When \( c = +\frac{\sqrt{65}}{2\sqrt{3}} \)
\( x_1 = \frac{-(5/4)(\sqrt{3})}{(\sqrt{65}/(2\sqrt{3}))} = \frac{-5\sqrt{3}}{4} \times \frac{2\sqrt{3}}{\sqrt{65}} = \frac{-5 \times 3 \times 2}{4\sqrt{65}} = \frac{-30}{4\sqrt{65}} = \frac{-15}{2\sqrt{65}} \)
\( y_1 = \frac{5/3}{(\sqrt{65}/(2\sqrt{3}))} = \frac{5}{3} \times \frac{2\sqrt{3}}{\sqrt{65}} = \frac{10\sqrt{3}}{3\sqrt{65}} \)

Case 2: When \( c = -\frac{\sqrt{65}}{2\sqrt{3}} \)
\( x_1 = \frac{-(5/4)(\sqrt{3})}{(-\sqrt{65}/(2\sqrt{3}))} = \frac{-5\sqrt{3}}{4} \times \frac{-2\sqrt{3}}{\sqrt{65}} = \frac{30}{4\sqrt{65}} = \frac{15}{2\sqrt{65}} \)
\( y_1 = \frac{5/3}{(-\sqrt{65}/(2\sqrt{3}))} = \frac{5}{3} \times \frac{-2\sqrt{3}}{\sqrt{65}} = \frac{-10\sqrt{3}}{3\sqrt{65}} \)

So, the coordinates of the tangent points are \( \left(\pm \frac{15}{2\sqrt{65}}, \mp \frac{10\sqrt{3}}{3\sqrt{65}}\right) \). The signs are opposite for x and y because of how c affects them. For instance, if x is positive, y is negative, and vice-versa.
In simple words: First, we converted the ellipse equation to a standard form. Then, we found the slope of the tangent line using the given angle. We used a special formula to write down the two possible equations for the tangent lines. After that, we used another formula to find the exact points where these lines touch the ellipse.

🎯 Exam Tip: When finding equations of tangents and tangent points, always convert the ellipse to its standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) first. Remember that the tangent point coordinates will have opposite signs for x and y corresponding to the \( \pm \) in the tangent equation.