Important Questions: Conic Sections | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: Compute the centre and radius of the circle 2x² + 2y² – x = 0
Ans: Given that, the circle equation is 2x² + 2y² – x = 0
This can be written as:
⇒ (2x²-x) + y² = 0
⇒ 2{[x² – (x/2)] +y2} = 0
⇒{ x² – 2x(¼) + (¼)²} +y² – (¼)² = 0
Now, simplify the above form, we get
⇒(x- (¼))² + (y-0)² = (¼)2
The above equation is of the form (x – h)² + (y – k)²= r²
Therefore, by comparing the general form and the equation obtained, we can say
h= ¼ , k = 0, and r = ¼.

Q2: Determine the foci coordinates, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse (x²/49) + (y²/36) = 1
Ans: The given equation is (x²/49) + (y²/36) = 1
It can be written as (x²/7²) + (y²/6²) = 1
It is noticed that the denominator of x²/49 is greater than the denominator of the y²/36
On comparing the equation with (x²/a²) + (y²/b²) = 1, we will get
a = 7 and b = 6
Therefore, c = √(a²– b²)
Now, substitute the value of a and b
⇒ √(a²– b²) = √(7²– 6²) = √(49-36)
⇒ √13
Hence, the foci coordinates are ( ± √13, 0)
Eccentricity, e = c/a = √13/ 7
Length of the major axis = 2a = 2(7) = 14
Length of the minor axis = 2b = 2(6) =12
The coordinates of the vertices are ( ± 7, 0)
Latus rectum Length= 2b²/a = 2(6)²/7 = 2(36)/7 = 72/7

Q3: Determine the equation of the hyperbola which satisfies the given conditions: Foci (0, ±13), the conjugate axis is of length 24.
Ans: Given that: Foci (0, ±13), Conjugate axis length = 24
It is noted that the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form:
(y²/α²)-(x²/b²) = 1 …(1)
Since the foci are (0, ±13), we can get
C = 13
It is given that, the length of the conjugate axis is 24,
It becomes 2b = 24
b = 24/2
b = 12
And, we know that α² + b² = c²
To find a, substitute the value of b and c in the above equation:
α² + 12² = 13²
α² = 169-144
α² = 25
Now, substitute the value of a and b in equation (1), we get
(y²/25)-(x²/144) = 1, which is the required equation of the hyperbola.

Q4: Determine the equation of the circle with radius 4 and Centre (-2, 3).
Ans: Given that:
Radius, r = 4, and center (h, k) = (-2, 3).
We know that the equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r² ….(1)
Now, substitute the radius and center values in (1), we get
Therefore, the equation of the circle is
(x + 2)² + (y – 3)² = (4)²
x² + 4x + 4 + y² – 6y + 9 = 16
Now, simplify the above equation, we get:
x² + y²+ 4x – 6y – 3 = 0
Thus, the equation of a circle with center (-2, 3) and radius 4 is x² + y² + 4x – 6y – 3 = 0

Q5: Determine the focus coordinates, the axis of the parabola, the equation of the directrix and the latus rectum length for y2 = -8x
Ans: Given that, the parabola equation is y2= -8x.
It is noted that the coefficient of x is negative.
Therefore, the parabola opens towards the left.
Now, compare the equation with y2= -4ax, we obtain
-4a = -8
⇒ a = 2
Thus, the value of a is 2.
Therefore, the coordinates of the focus = (-a, 0) = (-2, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = a i.e., x = 2
We know the formula to find the length of a latus rectum
Latus rectum length = 4a
Now, substitute a = 2, we get
Length of latus rectum = 8

Q6: Determine the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), the major axis on the y-axis and passes through the points (3, 2) and (1, 6).
Ans: Centre = (0, 0), and major axis that passes through the points (3, 2) and (1, 6).
We know that the equation of the ellipse will be of the form when the centre is at (0, 0) and the major axis is on the y-axis,
(x²/b²) + (y²/a²) = 1 …. (1)
Here, a is the semi-major axis.
It is given that, the ellipse passes through the points (3, 2) and (1, 6).
Hence, equation (1) becomes
(9/b²) + (4/a²) = 1 …(2)
(1/b²) + (36/a²) = 1 …(3)
Solving equation (2) and (3), we get
b² = 10 and a² = 40
Therefore, the equation of the ellipse becomes: (x²/10) + (y²/40) = 1