CBSE Class 11 Mathematics Conic Sections Assignment Set 02

Chapter 11

CONIC SECTION

Extra Questions:

Question 1. Find the centre and the radius of \(3x^2 + 3y^2 + 6x - 4y - 1 = 0\)
Answer: Given equation is \(3x^2 + 3y^2 + 6x - 4y - 1 = 0\)
Divide the entire equation by 3 to make the coefficients of \(x^2\) and \(y^2\) equal to 1:
\(x^2 + y^2 + 2x - \frac{4}{3}y - \frac{1}{3} = 0\)
Compare this with the general equation of a circle \(x^2 + y^2 + 2gx + 2fy + c = 0\):
\(2g = 2 \implies g = 1\)
\(2f = -\frac{4}{3} \implies f = -\frac{2}{3}\)
\(c = -\frac{1}{3}\)
Centre of the circle is given by \((-g, -f) = \left(-1, \frac{2}{3}\right)\)
Radius \(r = \sqrt{g^2 + f^2 - c}\)
\(r = \sqrt{(1)^2 + \left(-\frac{2}{3}\right)^2 - \left(-\frac{1}{3}\right)}\)
\(r = \sqrt{1 + \frac{4}{9} + \frac{1}{3}} = \sqrt{\frac{9 + 4 + 3}{9}} = \sqrt{\frac{16}{9}} = \frac{4}{3}\)
Centre = \(\left(-1, \frac{2}{3}\right)\), Radius = \(\frac{4}{3}\)
In simple words: By making the coefficients of squared terms equal to 1, we found the centre and radius using standard circle formulas.

Exam Tip: Always ensure the coefficients of \(x^2\) and \(y^2\) are 1 before applying the centre \((-g, -f)\) and radius \(\sqrt{g^2 + f^2 - c}\) formulas.

Question 2. Find the value of \(p\) so that \(x^2 + y^2 + 8x + 10y + p = 0\), is the equation of the circle of radius 7 units.
Answer: Given equation of the circle is \(x^2 + y^2 + 8x + 10y + p = 0\)
Comparing with the general equation \(x^2 + y^2 + 2gx + 2fy + c = 0\):
\(2g = 8 \implies g = 4\)
\(2f = 10 \implies f = 5\)
\(c = p\)
Given radius \(r = 7\)
We know that radius \(r = \sqrt{g^2 + f^2 - c}\)
\(7 = \sqrt{(4)^2 + (5)^2 - p}\)
Square both sides:
\(49 = 16 + 25 - p\)
\(49 = 41 - p\)
\(p = 41 - 49 = -8\)
In simple words: Substitute the given radius and coefficients into the radius formula to solve for the unknown parameter \(p\).

Exam Tip: Squaring both sides to eliminate the square root is a standard technique when finding unknown constants from a given radius.

Question 3. Find the equation of the circle when the end points of the diameter are \(A(-2, 3)\) and \(B(3, -5)\).
Answer: Let the endpoints of the diameter be \((x_1, y_1) = (-2, 3)\) and \((x_2, y_2) = (3, -5)\).
Using the diameter form of the equation of a circle:
\((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\)
Substitute the coordinates:
\((x - (-2))(x - 3) + (y - 3)(y - (-5)) = 0\)
\((x + 2)(x - 3) + (y - 3)(y + 5) = 0\)
Expand the terms:
\(x^2 - 3x + 2x - 6 + y^2 + 5y - 3y - 15 = 0\)
\(x^2 + y^2 - x + 2y - 21 = 0\)
In simple words: The diameter form equation directly gives the circle equation using its two endpoints.

Exam Tip: Memorize the diameter form equation \((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\) to save time during exams.

Question 4. Find the equation of the circle circumscribing the triangle formed by the straight lines: \(x + y = 6\), \(2x + y = 4\) and \(x + 2y = 5\)
Answer: First, find the vertices of the triangle by solving the lines pairwise:
Solving \(x + y = 6\) and \(2x + y = 4\):
Subtracting the first from the second gives \(x = -2\), then \(y = 8\). Vertex \(P(-2, 8)\).
Solving \(2x + y = 4\) and \(x + 2y = 5\):
Multiply first by 2: \(4x + 2y = 8\). Subtracting \(x + 2y = 5\) gives \(3x = 3 \implies x = 1\), then \(y = 2\). Vertex \(Q(1, 2)\).
Solving \(x + y = 6\) and \(x + 2y = 5\):
Subtracting gives \(y = -1\), then \(x = 7\). Vertex \(R(7, -1)\).
Let the general equation of the circle be \(x^2 + y^2 + 2gx + 2fy + c = 0\).
Substitute point \(P(-2, 8)\):
\((-2)^2 + (8)^2 + 2g(-2) + 2f(8) + c = 0 \implies 4 + 64 - 4g + 16f + c = 0 \implies -4g + 16f + c = -68\) (Equation 1)
Substitute point \(Q(1, 2)\):
\((1)^2 + (2)^2 + 2g(1) + 2f(2) + c = 0 \implies 1 + 4 + 2g + 4f + c = 0 \implies 2g + 4f + c = -5\) (Equation 2)
Substitute point \(R(7, -1)\):
\((7)^2 + (-1)^2 + 2g(7) + 2f(-1) + c = 0 \implies 49 + 1 + 14g - 2f + c = 0 \implies 14g - 2f + c = -50\) (Equation 3)
Subtract Equation 2 from Equation 1:
\(-6g + 12f = -63 \implies 2g - 4f = 21\) (Equation 4)
Subtract Equation 3 from Equation 2:
\(-12g + 6f = 45 \implies 4g - 2f = -15\) (Equation 5)
Multiply Equation 4 by 2: \(4g - 8f = 42\). Subtracting from Equation 5 gives:
\(6f = -57 \implies f = -\frac{19}{2}\)
Substitute \(f\) into Equation 4: \(2g - 4\left(-\frac{19}{2}\right) = 21 \implies 2g + 38 = 21 \implies 2g = -17 \implies g = -\frac{17}{2}\)
Substitute \(g\) and \(f\) into Equation 2:
\(2\left(-\frac{17}{2}\right) + 4\left(-\frac{19}{2}\right) + c = -5 \implies -17 - 38 + c = -5 \implies -55 + c = -5 \implies c = 50\)
Thus, the equation of the circle is \(x^2 + y^2 - 17x - 19y + 50 = 0\).
In simple words: Find the intersection points of the three lines to get the triangle vertices, then substitute them into the general circle equation to solve for constants.

Exam Tip: This is a lengthy problem; solve simultaneous linear equations carefully to avoid sign errors.

Question 5. Find the area of the triangle formed by the lines joining the vertex of the parabola \(x^2 = 12y\) to the ends of its latus rectum.
Answer: Given parabola is \(x^2 = 12y\), which is a vertical upward parabola of the form \(x^2 = 4ay\).
Here, \(4a = 12 \implies a = 3\).
Vertex of the parabola is \(O(0, 0)\).
The coordinates of the foci and latus rectum endpoints for \(x^2 = 4ay\) are \((-2a, a)\) and \((2a, a)\).
Substituting \(a = 3\), the ends of the latus rectum are \(L_1(-6, 3)\) and \(L_2(6, 3)\).
The triangle is formed by vertices \((0, 0)\), \((-6, 3)\), and \((6, 3)\).
Base of the triangle (distance between \(L_1\) and \(L_2\)) = \(6 - (-6) = 12\) units.
Height of the triangle (y-coordinate of the latus rectum) = \(3\) units.
Area of the triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 3 = 18\) sq. units.
In simple words: The base is the length of the latus rectum and the height is the distance from the vertex to the latus rectum line.

Exam Tip: For any standard parabola \(x^2 = 4ay\), the area of the triangle formed by the vertex and latus rectum endpoints is always \(\frac{1}{2} \times (4a) \times a = 2a^2\).

Question 6. Find the equation of the ellipse with eccentricity \(\frac{3}{4}\), foci on y-axis, center at the origin and passes through the point \((6, 4)\)
Answer: Since foci are on the y-axis and center is at the origin, the standard equation of the ellipse is:
\(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)
Given eccentricity \(e = \frac{3}{4}\), where \(c = ae \implies c = \frac{3}{4}a\).
We know the relation between semi-axes and focal distance: \(b^2 = a^2 - c^2\)
\(b^2 = a^2 - \left(\frac{3}{4}a\right)^2 = a^2 - \frac{9}{16}a^2 = \frac{7}{16}a^2\)
Substitute \(b^2\) into the ellipse equation:
\(\frac{x^2}{\frac{7}{16}a^2} + \frac{y^2}{a^2} = 1 \implies \frac{16x^2}{7a^2} + \frac{y^2}{a^2} = 1\)
Since the ellipse passes through \((6, 4)\), substitute \(x = 6\) and \(y = 4\):
\(\frac{16(6)^2}{7a^2} + \frac{(4)^2}{a^2} = 1\)
\(\frac{16(36)}{7a^2} + \frac{16}{a^2} = 1\)
\(\frac{576}{7a^2} + \frac{16}{a^2} = 1\)
Multiply the entire equation by \(a^2\):
\(\frac{576}{7} + 16 = a^2\)
\(a^2 = \frac{576 + 112}{7} = \frac{688}{7}\)
Now find \(b^2\):
\(b^2 = \frac{7}{16}a^2 = \frac{7}{16} \times \frac{688}{7} = \frac{688}{16} = 43\)
Substitute \(b^2 = 43\) and \(a^2 = \frac{688}{7}\) into the ellipse equation, or note that \(\frac{x^2}{43} + \frac{y^2}{\frac{688}{7}} = 1 \implies 16x^2 + 7y^2 = 688\).
In simple words: Use the vertical ellipse formula, substitute the eccentricity relation, and plug in the given point to find \(a^2\) and \(b^2\).

Exam Tip: Pay close attention to whether the foci are on the x-axis or y-axis to choose the correct standard ellipse form.

Question 7. Find the length of major axis and minor axis of \(4x^2 + y^2 = 100\)
Answer: Given equation is \(4x^2 + y^2 = 100\)
Divide by 100 to convert it into standard form:
\(\frac{4x^2}{100} + \frac{y^2}{100} = 1 \implies \frac{x^2}{25} + \frac{y^2}{100} = 1\)
Here, the denominator under \(y^2\) is greater, so \(a^2 = 100 \implies a = 10\) and \(b^2 = 25 \implies b = 5\).
Length of major axis = \(2a = 2(10) = 20\) units.
Length of minor axis = \(2b = 2(5) = 10\) units.
In simple words: Rewrite the equation in standard form by dividing by the constant on the right side, then identify \(a\) and \(b\).

Exam Tip: The major axis is always associated with the larger denominator in the ellipse equation.

Question 8. Find the equation of the parabola with the centre at origin, length of transverse axis 6 units and a focus at \((0, 4)\).
Answer: Note: The question refers to a hyperbola (as indicated by "transverse axis" and the answer key format).
Since the focus is at \((0, 4)\), which lies on the y-axis, it is a vertical hyperbola of the form:
\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)
Length of transverse axis = \(2a = 6 \implies a = 3 \implies a^2 = 9\).
Focus is given as \((0, c) = (0, 4) \implies c = 4 \implies c^2 = 16\).
We know that for a hyperbola, \(b^2 = c^2 - a^2\):
\(b^2 = 16 - 9 = 7\)
Substitute \(a^2 = 9\) and \(b^2 = 7\) into the hyperbola equation:
\(\frac{y^2}{9} - \frac{x^2}{7} = 1\)
Multiply by 63 to clear denominators:
\(7y^2 - 9x^2 = 63\)
In simple words: Transverse axis length gives \(a\), the focus gives \(c\), and we use the hyperbola relation \(b^2 = c^2 - a^2\) to find \(b^2\).

Exam Tip: Transverse axis belongs to hyperbolas, while major axis belongs to ellipses.

Question 9. The line \(5x - y = 3\) is a tangent to a circle at a point \((2, 7)\) and its centre is on the line \(x + 2y = 19\). Find the equation of the circle.
Answer: Let the centre of the circle be \((h, k)\).
Since the centre lies on the line \(x + 2y = 19\):
\(h + 2k = 19 \implies h = 19 - 2k\) (Equation 1)
The radius is the distance from the centre \((h, k)\) to the point of tangency \((2, 7)\):
\(r^2 = (h - 2)^2 + (k - 7)^2\)
Also, the radius is perpendicular to the tangent line \(5x - y - 3 = 0\). The slope of the tangent is \(m_1 = 5\).
The slope of the line joining centre \((h, k)\) and point \((2, 7)\) is \(m_2 = \frac{k - 7}{h - 2}\).
Since radius is perpendicular to tangent, \(m_1 \times m_2 = -1\):
\(5 \times \left(\frac{k - 7}{h - 2}\right) = -1 \implies 5k - 35 = -h + 2 \implies h + 5k = 37\) (Equation 2)
Substitute Equation 1 into Equation 2:
\((19 - 2k) + 5k = 37\)
\(19 + 3k = 37\)
\(3k = 18 \implies k = 6\)
Substitute \(k = 6\) into Equation 1:
\(h = 19 - 2(6) = 19 - 12 = 7\)
So, the centre of the circle is \((h, k) = (7, 6)\).
Now, find the radius squared \(r^2\):
\(r^2 = (7 - 2)^2 + (6 - 7)^2 = (5)^2 + (-1)^2 = 25 + 1 = 26\)
Using the standard equation of the circle with centre \((7, 6)\) and radius squared \(26\):
\((x - 7)^2 + (y - 6)^2 = 26\)
\(x^2 - 14x + 49 + y^2 - 12y + 36 = 26\)
\(x^2 + y^2 - 14x - 12y + 59 = 0\)
In simple words: Use the perpendicularity between the radius and tangent to find the centre coordinates, then calculate the radius.

Exam Tip: The product of slopes of two perpendicular lines is always \(-1\), which is key for finding the centre when a tangent is given.

Question 10. Find equation of the circle which touches the y-axis at origin and whose radius is 3 units.
Answer: Since the circle touches the y-axis at the origin \((0, 0)\), its centre must lie on the x-axis.
Since the radius is 3 units and it touches the y-axis at origin, the centre is located at \((3, 0)\) or \((-3, 0)\).
Case 1: When centre is \((3, 0)\) and radius \(r = 3\):
Equation is \((x - 3)^2 + (y - 0)^2 = 3^2\)
\(x^2 - 6x + 9 + y^2 = 9 \implies x^2 + y^2 - 6x = 0\)
Case 2: When centre is \((-3, 0)\) and radius \(r = 3\):
Equation is \((x + 3)^2 + (y - 0)^2 = 3^2\)
\(x^2 + 6x + 9 + y^2 = 9 \implies x^2 + y^2 + 6x = 0\)
Matching with the given answer key, the equation is \(x^2 + y^2 - 6x = 0\).
In simple words: A circle touching the y-axis at the origin has its centre on the x-axis at a distance equal to its radius.

Exam Tip: Whenever a circle touches a coordinate axis at the origin, its centre coordinate along that axis equals the radius in magnitude.