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Q1. The equation of parabola whose focus is (– 3, 0) and directrix x + 5 = 0 is:

• y² = 4x
• y² = 4(x + 1)
• y² = 4(x + 16)
• y² = 4(x + 4)

Solution

The required equation of parabola is:

Q2. Equation of circle whose centre is origin and radius is 3 is:

• (x – h)² + (y – k)² = 3
• (x – h)² + (y – k)² = 9
• x² + y² = 9
• x² + y² = 3

Solution

Equation of circle is: (x - 0)² + (y – 0)² = (3)² or x² + y² = 9

Q3. A circle is the set of …… in a plane that are equidistant from a fixed point in the plane.

• Lines
• Circles
• Points
• Cones

Solution

A circle is the set of points in a plane that are equidistant from a fixed point in the plane. The fixed point is called the centre of the circle and the distance from the centre to a point on the circle is called the radius of the circle.

Q4. The coordinates of the focus and length of latus-rectum of the parabola y² = 8x is:

• (2, 0), 8
• (4, 0), 8
• (4, 0), 16
• (2, 0), 16

Solution

Comparing with the given equation y² = 4ax, we find that a = 2. Thus, the focus of the parabola is (2, 0) and length of the latus-rectum is 4a = 4 × 2 = 8.

Q5. The equation of the parabola with vertex at origin, which passes through the point (– 3, 7) and axis along the X-axis is:

• y² = 49x
• 3y² = – 49x
• 3y² = 49x
• x² = – 49y

Solution

Let a parabola with vertex at origin and axis along the X-axis be y² = 4ax. It passes through (–3, 7).  The required equation of the parabola is  or 3y² = – 49x

Q6. The …… of a conic is the chord passing through the focus and perpendicular to the axis.

• Line of axis
• Y-axis
• Latus-rectum
• Directrix

Solution

The latus-rectum of a conic is the chord passing through the focus and perpendicular to the axis.

Q7. The foci of the ellipse 25 (x + 1)² + 9(y + 2)² = 225 are at:

• (–1, 2) and (–1, –6)
• (–2, 1) and (–2, 6)
• (–1, –2) and (–2, –1)
• (–1, –2) and (–1, –6)

Solution

The given eq. is Centre of the ellipse is (–1, –2) and a = 3, b = 5, so that a < b. For and ellipse, Hence, foci are , i.e., foci are (–1, –6) and (–1, 2).

Q8. A parabola has its axis of symmetry as the x axis. The parabola opens to the left, so its  equation is of the type

• x²=4ay , a>0
• x²= -4ay , a>0
• y² =4ax , a>0
• y² = - 4ax , a>0

Solution

For a parabola which has x axis as the axis of symmetry, the equation has to be of the type _y² =4ax , a>0 or _y² = - 4ax , a>0.Moreover, if it opens to the left, then that clears the way and we get the answer _y² = - 4ax , a>0.

Q9. The parabolax²= - 4ay

• Opens to the left
• Opens to the right
• Opens upwards
• Opens downwards

Solution

A parabola with this type of equation has its focus at (0,-a) and vertex at (0,0,). This fact leads us to the answer.

Q10. The equation 2x² - 3xy + 5y² + 6x - 3y + 5 = 0 represents.

• A parabola
• An ellipse
• A hyperbola
• A pair of straight lines

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