Class XI Chapter **11** – Conic Section Maths ______________________________________________________________________________ Question 9: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0) Solution 9: Vertex (0, 0); focus (3, 0) Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y 2 = 4ax. Since the focus is (3, 0), a= 3. Thus, the equation of the parabola is y 2 = 4 x 3 x x, i.e., y 2 = 12x Question 10: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (-2, 0) Solution 10: Vertex (0,0) focus (-2, 0) Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y 2 = -4ax. Since the focus is (-2, 0), a = 2. Thus, the equation of the parabola is y 2 = -4(2)x, i.e., y 2 = -8x Question **11**: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis Solution **11**: Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form y 2 = 4ax or y2= â€“4ax. The parabola passes through point (2, 3), which lies in the first quadrant. Therefore, the equation of the parabola is of the form y 2 = 4ax,while point (2, 3) must satisfy the equation y 2 = 4ax. 3 4a 2 a 8 Thus, the equation of the parabola is 2 9 y 4 x 8 2 9 y x 2 2 2y 9x 2 9 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter **11** – Conic Section Maths ______________________________________________________________________________ Question 12: Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis Solution 12: Since the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form x 2 = 4ay or x 2 = â€“4ay. The parabola passes through point (5, 2), which lies in the first quadrant. Therefore, the equation of the parabola is of the form x 2 = 4ay,while point (5, 2) must satisfy the equation x 2 = 4ay. 5 2 4 a 2 25 8a a 25 8 Thus, the equation of the parabola is 2 25 x 4 y 8 2 2x 25y Exercise **11**.3 Question 1: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse Solution 1: 2 2 x y 1 36 16 2 2 x y The given equation is 1 36 16 2 2 x y Here, the denominator of is greater than the denominator of . 36 16 Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis 2 2 x y On comparing the given equation with 1 , we obtain a = 6 and b = 4. 2 2 a b 2 2 c a b 36 16 20 2 5 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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