Class - 11, Ex - 10.2 Q1 to Q12 Conic Sections Maths || CBSE NCERT || New Book || Green Board

Overview

The video solves all 12 questions in Class 11 NCERT Exercise 10.2 on conic sections, focusing on parabolas. Questions 1–6 ask for a parabola’s focus, axis, directrix, and latus rectum. Questions 7–12 ask for the parabola’s equation from given information.

Key Formulas and Method

For a parabola with its vertex at the origin:

Direction of opening Standard form Focus Axis Directrix Right (y^2=4ax) ((a,0)) (x)-axis (x=-a) Left (y^2=-4ax) ((-a,0)) (x)-axis (x=a) Upward (x^2=4ay) ((0,a)) (y)-axis (y=-a) Downward (x^2=-4ay) ((0,-a)) (y)-axis (y=a)

In each form, the length of the latus rectum is (4a). To solve, compare the given equation with the appropriate standard form to find (a), then use the corresponding formulas for the focus, axis, and directrix. When an equation is requested, use the given focus, directrix, or point on the parabola to determine the correct form and value of (a).

Questions 1–6: Find the Focus, Axis, Directrix, and Latus Rectum

Question Given equation Focus Axis Directrix Latus rectum 1 (y^2=12x) ((3,0)) (x)-axis (x=-3) (12) 2 (x^2=6y) ((0,\frac32)) (y)-axis (y=-\frac32) (6) 3 (y^2=-8x) ((-2,0)) (x)-axis (x=2) (8) 4 (x^2=-16y) ((0,-4)) (y)-axis (y=4) (16) 5 (y^2=10x) ((\frac52,0)) (x)-axis (x=-\frac52) (10) 6 (x^2=-9y) ((0,-\frac94)) (y)-axis (y=\frac94) (9)

The opening direction follows from the sign of the coefficient: a positive coefficient gives rightward or upward opening, while a negative coefficient gives leftward or downward opening.

Questions 7–12: Find the Parabola’s Equation

The solutions use the standard forms above, with the vertex at the origin.

• Question 7: The directrix is (x=-6), so the parabola opens right and (a=6): [ y^2=24x. ]

• Question 8: The directrix is (y=3), so the parabola opens downward and (a=3): [ x^2=-12y. ]

• Question 9: The focus is ((3,0)), so the parabola opens right with (a=3): [ y^2=12x. ]

• Question 10: The focus is ((-2,0)), so the parabola opens left with (a=2): [ y^2=-8x. ]

• Question 11: The parabola has its axis along the (x)-axis and passes through ((2,3)). Substituting into (y^2=4ax) gives (a=\frac98): [ y^2=\frac92x, ] or [ 2y^2-9x=0. ]

• Question 12: The parabola is symmetric about the (y)-axis and passes through ((5,2)). Substituting into (x^2=4ay) gives (a=\frac{25}{8}): [ x^2=\frac{25}{2}y, ] or [ 2x^2-25y=0. ]

Speaker and Source

• Mandeep — the instructor presenting the lesson on the Green Board Class 11th and 12th channel.