RBSE Solutions Class 11 Maths Chapter 12 Conic Section Exercise 12.5

Question 1. Find the equation of ellipse whose :
(i) Focus (- 1, 1), Directrix x - y + 4 = 0 and eccentricity is e = \( \frac { 1 }{ \sqrt{5} } \)
(ii) Focus (- 2,3), Directrix 3x + 4y = 1 and eccentricity is e = \( \frac { 1 }{ 3 } \)
Answer:
(i) Let \( P(h, k) \) be any point on the ellipse. By definition, the distance from \( P \) to the focus is \( e \) times the distance from \( P \) to the directrix.
So, \( PS = e(PM) \)
Squaring both sides gives: \( (PS)^2 = e^2(PM)^2 \)
The focus is \( (-1, 1) \), the directrix is \( x - y + 4 = 0 \), and eccentricity \( e = \frac { 1 }{ \sqrt{5} } \).
The formula for distance \( PS \) is \( \sqrt{ (h - (-1))^2 + (k - 1)^2 } \).
The formula for distance \( PM \) is \( \frac{ |h - k + 4|}{ \sqrt{1^2 + (-1)^2} } \).
Substitute these into the squared equation:
\( (h + 1)^2 + (k - 1)^2 = \left( \frac{1}{\sqrt{5}} \right)^2 \left( \frac{h - k + 4}{\sqrt{1^2 + (-1)^2}} \right)^2 \)
\( (h + 1)^2 + (k - 1)^2 = \frac{1}{5} \left( \frac{(h - k + 4)^2}{2} \right) \)
Multiply both sides by 10 to remove the denominators:
\( 10 [ (h + 1)^2 + (k - 1)^2 ] = (h - k + 4)^2 \)
Expand the terms:
\( 10 [ (h^2 + 2h + 1) + (k^2 - 2k + 1) ] = h^2 + k^2 + 16 - 2hk + 8h - 8k \)
\( 10h^2 + 20h + 10 + 10k^2 - 20k + 10 = h^2 + k^2 - 2hk + 8h - 8k + 16 \)
Move all terms to one side:
\( (10h^2 - h^2) + (10k^2 - k^2) + 2hk + (20h - 8h) + (-20k + 8k) + (10 + 10 - 16) = 0 \)
\( 9h^2 + 9k^2 + 2hk + 12h - 12k + 4 = 0 \)
Finally, replace \( h \) with \( x \) and \( k \) with \( y \) to find the locus of point \( P \):
\( 9x^2 + 9y^2 + 2xy + 12x - 12y + 4 = 0 \)
This is the required equation of the ellipse.
(ii) Let \( P(h, k) \) be any point on the ellipse. As per the definition, the distance from \( P \) to the focus is \( e \) times the distance from \( P \) to the directrix.
So, \( PS = e(PM) \)
Squaring both sides gives: \( (PS)^2 = e^2(PM)^2 \)
The focus is \( (-2, 3) \), the directrix is \( 3x + 4y - 1 = 0 \), and eccentricity \( e = \frac { 1 }{ 3 } \).
The distance \( PS \) is \( \sqrt{ (h - (-2))^2 + (k - 3)^2 } \).
The distance \( PM \) is \( \frac{ |3h + 4k - 1|}{ \sqrt{3^2 + 4^2} } \).
Substitute these into the squared equation:
\( (h + 2)^2 + (k - 3)^2 = \left( \frac{1}{3} \right)^2 \left( \frac{3h + 4k - 1}{\sqrt{3^2 + 4^2}} \right)^2 \)
\( (h + 2)^2 + (k - 3)^2 = \frac{1}{9} \left( \frac{(3h + 4k - 1)^2}{25} \right) \)
Multiply both sides by \( 9 \times 25 = 225 \) to remove the denominators:
\( 225 [ (h + 2)^2 + (k - 3)^2 ] = (3h + 4k - 1)^2 \)
Expand the terms:
\( 225 [ (h^2 + 4h + 4) + (k^2 - 6k + 9) ] = (9h^2 + 16k^2 + 1 + 24hk - 6h - 8k) \)
\( 225h^2 + 900h + 900 + 225k^2 - 1350k + 2025 = 9h^2 + 16k^2 + 1 + 24hk - 6h - 8k \)
Move all terms to one side and combine like terms:
\( (225h^2 - 9h^2) + (225k^2 - 16k^2) - 24hk + (900h + 6h) + (-1350k + 8k) + (900 + 2025 - 1) = 0 \)
\( 216h^2 + 209k^2 - 24hk + 906h - 1342k + 2924 = 0 \)
Finally, replace \( h \) with \( x \) and \( k \) with \( y \) to find the locus of point \( P \):
\( 216x^2 + 209y^2 - 24xy + 906x - 1342y + 2924 = 0 \)
This is the required equation of the ellipse.
In simple words: To find the ellipse equation, we use a rule: the distance from any point on the ellipse to its special point (focus) is always a fixed fraction (eccentricity) of its distance to a special line (directrix). We use this rule and algebra to get the final equation.

🎯 Exam Tip: Remember the fundamental definition of an ellipse as the locus of a point whose distance from a fixed point (focus) bears a constant ratio (eccentricity) to its distance from a fixed line (directrix). This is key for deriving the equation.

Question 2. Find the eccentricity, latus rectum and focus of the following ellipse :
(i) \( 4x^2 + 9y^2 = 1 \)
(ii) \( 25x^2 + 4y^2 = 100 \)
(iii) \( 3x^2 + 4y^2 - 12x - 8y + 4 = 0 \)
Answer:
(i) Given the equation of the ellipse: \( 4x^2 + 9y^2 = 1 \).
To get the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can rewrite it as:
\( \frac{x^2}{1/4} + \frac{y^2}{1/9} = 1 \)
From this, we can see that \( a^2 = \frac{1}{4} \) so \( a = \frac{1}{2} \).
And \( b^2 = \frac{1}{9} \) so \( b = \frac{1}{3} \).
Since \( a > b \), the major axis is along the x-axis.
To find the eccentricity \( e \), we use the formula \( b^2 = a^2(1 - e^2) \).
Substitute the values of \( a^2 \) and \( b^2 \):
\( \frac{1}{9} = \frac{1}{4}(1 - e^2) \)
Multiply both sides by 4:
\( \frac{4}{9} = 1 - e^2 \)
Now, solve for \( e^2 \):
\( e^2 = 1 - \frac{4}{9} \)
\( e^2 = \frac{9 - 4}{9} \)
\( e^2 = \frac{5}{9} \)
Take the square root to find \( e \):
\( e = \frac{\sqrt{5}}{3} \)
Next, calculate the length of the latus rectum using the formula \( \frac{2b^2}{a} \).
\( \text{Latus rectum} = \frac{2 \times (1/9)}{1/2} = \frac{2/9}{1/2} = \frac{2}{9} \times 2 = \frac{4}{9} \)
Finally, find the coordinates of the foci, which are \( (\pm ae, 0) \).
\( ae = \frac{1}{2} \times \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{6} \)
The foci are \( \left( \frac{\sqrt{5}}{6}, 0 \right) \) and \( \left( -\frac{\sqrt{5}}{6}, 0 \right) \).
(ii) Given the equation of the ellipse: \( 25x^2 + 4y^2 = 100 \).
To convert it to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), divide the entire equation by 100:
\( \frac{25x^2}{100} + \frac{4y^2}{100} = \frac{100}{100} \)
\( \frac{x^2}{4} + \frac{y^2}{25} = 1 \)
From this, we can see that \( a^2 = 4 \) so \( a = 2 \).
And \( b^2 = 25 \) so \( b = 5 \).
Since \( b > a \), the major axis is along the y-axis.
To find the eccentricity \( e \), we use the formula \( a^2 = b^2(1 - e^2) \).
Substitute the values of \( a^2 \) and \( b^2 \):
\( 4 = 25(1 - e^2) \)
Divide by 25:
\( \frac{4}{25} = 1 - e^2 \)
Now, solve for \( e^2 \):
\( e^2 = 1 - \frac{4}{25} \)
\( e^2 = \frac{25 - 4}{25} \)
\( e^2 = \frac{21}{25} \)
Take the square root to find \( e \):
\( e = \frac{\sqrt{21}}{5} \)
Next, calculate the length of the latus rectum using the formula \( \frac{2a^2}{b} \).
\( \text{Latus rectum} = \frac{2 \times 4}{5} = \frac{8}{5} \)
Finally, find the coordinates of the foci, which are \( (0, \pm be) \).
\( be = 5 \times \frac{\sqrt{21}}{5} = \sqrt{21} \)
The foci are \( (0, \sqrt{21}) \) and \( (0, -\sqrt{21}) \).
(iii) Given the equation of the ellipse: \( 3x^2 + 4y^2 - 12x - 8y + 4 = 0 \).
First, group the \( x \) terms and \( y \) terms, and move the constant to the other side:
\( (3x^2 - 12x) + (4y^2 - 8y) = -4 \)
Factor out the coefficients of \( x^2 \) and \( y^2 \):
\( 3(x^2 - 4x) + 4(y^2 - 2y) = -4 \)
Now, complete the square for both the \( x \) and \( y \) expressions. For \( x^2 - 4x \), add \( (-4/2)^2 = 4 \). For \( y^2 - 2y \), add \( (-2/2)^2 = 1 \).
Remember to multiply the added values by their factored coefficients on the left side and add them to the right side.
\( 3(x^2 - 4x + 4) + 4(y^2 - 2y + 1) = -4 + 3(4) + 4(1) \)
\( 3(x - 2)^2 + 4(y - 1)^2 = -4 + 12 + 4 \)
\( 3(x - 2)^2 + 4(y - 1)^2 = 12 \)
To get the standard form \( \frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1 \), divide the entire equation by 12:
\( \frac{3(x - 2)^2}{12} + \frac{4(y - 1)^2}{12} = \frac{12}{12} \)
\( \frac{(x - 2)^2}{4} + \frac{(y - 1)^2}{3} = 1 \)
Let \( X = x - 2 \) and \( Y = y - 1 \). The equation becomes \( \frac{X^2}{4} + \frac{Y^2}{3} = 1 \).
From this, \( a^2 = 4 \) so \( a = 2 \).
And \( b^2 = 3 \) so \( b = \sqrt{3} \).
Since \( a > b \), the major axis is parallel to the x-axis.
To find the eccentricity \( e \), use the formula \( b^2 = a^2(1 - e^2) \).
Substitute the values:
\( 3 = 4(1 - e^2) \)
\( \frac{3}{4} = 1 - e^2 \)
\( e^2 = 1 - \frac{3}{4} \)
\( e^2 = \frac{1}{4} \)
Take the square root to find \( e \):
\( e = \frac{1}{2} \)
Next, calculate the length of the latus rectum using the formula \( \frac{2b^2}{a} \).
\( \text{Latus rectum} = \frac{2 \times 3}{2} = 3 \)
Finally, find the coordinates of the foci, which are \( (\pm ae, 0) \) in the \( (X, Y) \) coordinate system.
\( ae = 2 \times \frac{1}{2} = 1 \)
So, the foci are \( (X = \pm 1, Y = 0) \).
Now, convert back to \( (x, y) \) coordinates:
If \( X = 1 \), then \( x - 2 = 1 \implies x = 3 \). If \( Y = 0 \), then \( y - 1 = 0 \implies y = 1 \). So, one focus is \( (3, 1) \).
If \( X = -1 \), then \( x - 2 = -1 \implies x = 1 \). If \( Y = 0 \), then \( y - 1 = 0 \implies y = 1 \). So, the other focus is \( (1, 1) \).
Thus, the coordinates of the foci are \( (3, 1) \) and \( (1, 1) \).
In simple words: We find three key features of an ellipse: how "stretched out" it is (eccentricity), the length of a special chord through its focus (latus rectum), and the exact locations of its two special points (foci). We do this by changing the ellipse equation into a standard form and then using specific formulas.

🎯 Exam Tip: Always convert the given ellipse equation into its standard form before calculating eccentricity, latus rectum, and foci. Pay close attention to whether \( a > b \) or \( b > a \) to correctly identify the major axis and corresponding formulas.

Question 3. Find the equation of ellipse whose axis are coordinate axis and passes through points (6, 2) and (4, 3).
Answer:
We start with the standard equation of an ellipse whose axes are the coordinate axes: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
The ellipse passes through the point \( (6, 2) \). We substitute these coordinates into the equation:
\( \frac{6^2}{a^2} + \frac{2^2}{b^2} = 1 \)
\( \frac{36}{a^2} + \frac{4}{b^2} = 1 \) (Equation 1)
The ellipse also passes through the point \( (4, 3) \). We substitute these coordinates as well:
\( \frac{4^2}{a^2} + \frac{3^2}{b^2} = 1 \)
\( \frac{16}{a^2} + \frac{9}{b^2} = 1 \) (Equation 2)
To solve for \( a^2 \) and \( b^2 \), we can treat \( \frac{1}{a^2} \) and \( \frac{1}{b^2} \) as variables. Let \( A = \frac{1}{a^2} \) and \( B = \frac{1}{b^2} \).
So, the equations become:
\( 36A + 4B = 1 \) (1)
\( 16A + 9B = 1 \) (2)
Multiply Equation (1) by 9 and Equation (2) by 4 to eliminate \( B \):
\( 9 \times (36A + 4B) = 9 \times 1 \implies 324A + 36B = 9 \)
\( 4 \times (16A + 9B) = 4 \times 1 \implies 64A + 36B = 4 \)
Now, subtract the second new equation from the first:
\( (324A + 36B) - (64A + 36B) = 9 - 4 \)
\( 260A = 5 \)
Solve for \( A \):
\( A = \frac{5}{260} \)
\( A = \frac{1}{52} \)
Since \( A = \frac{1}{a^2} \), we have \( \frac{1}{a^2} = \frac{1}{52} \), which means \( a^2 = 52 \).
Now substitute the value of \( A \) back into Equation (1) to find \( B \):
\( 36 \left( \frac{1}{52} \right) + 4B = 1 \)
\( \frac{36}{52} + 4B = 1 \)
\( \frac{9}{13} + 4B = 1 \)
\( 4B = 1 - \frac{9}{13} \)
\( 4B = \frac{13 - 9}{13} \)
\( 4B = \frac{4}{13} \)
Solve for \( B \):
\( B = \frac{4}{13 \times 4} \)
\( B = \frac{1}{13} \)
Since \( B = \frac{1}{b^2} \), we have \( \frac{1}{b^2} = \frac{1}{13} \), which means \( b^2 = 13 \).
Therefore, the equation of the ellipse is \( \frac{x^2}{52} + \frac{y^2}{13} = 1 \).
In simple words: We used the standard ellipse equation and the two points it passes through to create two algebraic equations. By solving these two equations, we found the values for \( a^2 \) and \( b^2 \), which then gave us the full equation of the ellipse.

🎯 Exam Tip: When given points that an ellipse passes through, substitute them into the general equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) to form a system of linear equations in terms of \( \frac{1}{a^2} \) and \( \frac{1}{b^2} \). Solving this system will yield the required equation.

Question 4. Find the eccentricity of ellipse whose latus rectum is half of its minor axis.
Answer:
Let the standard equation of the ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. For this problem, we will assume \( a > b \).
The formula for the length of the latus rectum of an ellipse is \( \frac{2b^2}{a} \).
The length of the minor axis is \( 2b \).
According to the problem statement, the latus rectum is half of the minor axis.
So, we can write the equation:
\( \frac{2b^2}{a} = \frac{1}{2} (2b) \)
Simplify the right side of the equation:
\( \frac{2b^2}{a} = b \)
Since \( b \) cannot be zero (otherwise it wouldn't be an ellipse), we can divide both sides by \( b \):
\( \frac{2b}{a} = 1 \)
Multiply by \( a \) to solve for \( a \):
\( 2b = a \)
Now, we use the relationship between \( a \), \( b \), and the eccentricity \( e \):
\( b^2 = a^2(1 - e^2) \)
Substitute \( a = 2b \) into this equation:
\( b^2 = (2b)^2(1 - e^2) \)
\( b^2 = 4b^2(1 - e^2) \)
Since \( b^2 \) cannot be zero, we can divide both sides by \( b^2 \):
\( 1 = 4(1 - e^2) \)
Divide by 4:
\( \frac{1}{4} = 1 - e^2 \)
Now, solve for \( e^2 \):
\( e^2 = 1 - \frac{1}{4} \)
\( e^2 = \frac{4 - 1}{4} \)
\( e^2 = \frac{3}{4} \)
Take the square root to find the eccentricity \( e \):
\( e = \sqrt{\frac{3}{4}} \)
\( e = \frac{\sqrt{3}}{2} \)
Thus, the eccentricity of the ellipse is \( \frac{\sqrt{3}}{2} \).
In simple words: We used the given fact that the ellipse's latus rectum is half its minor axis. This helped us find a connection between the lengths of the axes. Then, using the standard formula that links these axis lengths to the eccentricity, we calculated the ellipse's eccentricity.

🎯 Exam Tip: Always remember the standard formulas for latus rectum and minor axis of an ellipse. The key to solving this type of problem is setting up the correct relationship based on the given condition and then using the standard eccentricity formula.