Find the coordinates of the focus, axis, equation of the directrix, and length of the latus rectum of the parabola y^2=12x.
Hint. Compare with y^2=4ax to find a; the axis is the x-axis since y is squared, and the coefficient is positive so it opens right.
Comparing y^2=12x with y^2=4ax gives 4a=12, so a=3.
✦ Working through each part gives: Focus (3,0); axis: the x-axis; directrix x=-3; latus rectum = 4a = 12.
Find the coordinates of the focus, axis, equation of the directrix, and length of the latus rectum of the parabola x^2=6y.
Hint. Compare with x^2=4ay to find a; the axis is the y-axis since x is squared.
Comparing x^2=6y with x^2=4ay gives 4a=6, so a=3/2.
✦ Working through each part gives: Focus (0,3/2); axis: the y-axis; directrix y=-3/2; latus rectum = 4a = 6.
Find the coordinates of the focus, axis, equation of the directrix, and length of the latus rectum of the parabola y^2=-8x.
Hint. Compare with y^2=-4ax; the negative sign means the parabola opens left.
Comparing y^2=-8x with y^2=-4ax gives 4a=8, so a=2.
✦ Working through each part gives: Focus (-2,0); axis: the x-axis; directrix x=2; latus rectum = 4a = 8.
Find the coordinates of the focus, axis, equation of the directrix, and length of the latus rectum of the parabola x^2=-16y.
Hint. Compare with x^2=-4ay; the negative sign means the parabola opens downward.
Comparing x^2=-16y with x^2=-4ay gives 4a=16, so a=4.
✦ Working through each part gives: Focus (0,-4); axis: the y-axis; directrix y=4; latus rectum = 4a = 16.
Find the coordinates of the focus, axis, equation of the directrix, and length of the latus rectum of the parabola y^2=10x.
Hint. Compare with y^2=4ax to find a.
Comparing y^2=10x with y^2=4ax gives 4a=10, so a=5/2.
✦ Working through each part gives: Focus (5/2,0); axis: the x-axis; directrix x=-5/2; latus rectum = 4a = 10.
Find the coordinates of the focus, axis, equation of the directrix, and length of the latus rectum of the parabola x^2=-9y.
Hint. Compare with x^2=-4ay to find a.
Comparing x^2=-9y with x^2=-4ay gives 4a=9, so a=9/4.
✦ Working through each part gives: Focus (0,-9/4); axis: the y-axis; directrix y=9/4; latus rectum = 4a = 9.
Find the equation of the parabola that satisfies: focus (6,0), directrix x=-6.
Hint. A focus on the positive x-axis with a directrix on the opposite side means the parabola opens right, of the form y^2=4ax with a equal to the focus's distance from the origin.
Since the focus (6,0) lies on the x-axis with directrix x=-6 on the opposite side of the origin, a=6 and the parabola opens right, so the equation is y^2=4(6)x.
✦ Working through each part gives: y^2=24x.
Find the equation of the parabola that satisfies: focus (0,-3), directrix y=3.
Hint. A focus below the origin with directrix above it means the parabola opens downward, of the form x^2=-4ay.
Since the focus (0,-3) lies on the y-axis with directrix y=3 on the opposite side, a=3 and the parabola opens downward, so the equation is x^2=-4(3)y.
✦ Working through each part gives: x^2=-12y.
Find the equation of the parabola that satisfies: vertex (0,0), focus (3,0).
Hint. The vertex is at the origin and the focus lies on the positive x-axis, so this is y^2=4ax with a as the vertex-to-focus distance.
Since the vertex is (0,0) and the focus (3,0) lies on the positive x-axis, a=3, giving y^2=4(3)x.
✦ Working through each part gives: y^2=12x.
Find the equation of the parabola that satisfies: vertex (0,0), focus (-2,0).
Hint. The focus lies on the negative x-axis, so the parabola opens left: y^2=-4ax.
Since the vertex is (0,0) and the focus (-2,0) lies on the negative x-axis, a=2, giving y^2=-4(2)x.
✦ Working through each part gives: y^2=-8x.
Find the equation of the parabola with vertex (0,0), passing through (2,3), with axis along the x-axis.
Hint. Since the axis is the x-axis and the point has a positive y-coordinate reachable either way, use y^2=4ax and substitute the given point to find a.
The equation has the form y^2=4ax. Substituting (2,3): 9=4a(2)=8a, so a=9/8, giving y^2=4(9/8)x.
✦ Working through each part gives: y^2=(9/2)x.
Find the equation of the parabola with vertex (0,0), passing through (5,2), symmetric about the y-axis.
Hint. Symmetric about the y-axis means the form x^2=4ay (or -4ay); use the given point to find a and its sign.
The equation has the form x^2=4ay. Substituting (5,2): 25=4a(2)=8a, so a=25/8, giving x^2=4(25/8)y.
✦ Working through each part gives: x^2=(25/2)y.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh110.pdf) — Exercise 10.1 (15 questions), Exercise 10.2 (12 questions), Exercise 10.3 (20 questions), Exercise 10.4 (15 questions), plus the chapter's Miscellaneous Exercise (8 questions), 70 questions total. Confirmed against the CBSE curriculum PDF that Conic Sections has no formative-only carve-out at all, unlike its neighbouring chapters Straight Lines and Introduction to Three-Dimensional Geometry — every phrase in its syllabus line is summative. Confirmed the book teaches the general circle equation only through completing the square (Example 3), never naming g, f, c as constants, and never mentions hyperbola asymptotes anywhere (zero hits for the word itself, and none of Exercise 10.4's 15 questions ask for one). Exercise 10.1 Q3's centre coordinates were cross-checked via a 300dpi page render after the raw PDF text extraction reversed their order. Several exercise answers (10.1 Q10/Q11, 10.4 Q15, Misc Q2/Q3/Q5/Q8) were cross-verified against known standard results for this exact problem set.. Questions are referenced from the NCERT textbook for identification.
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