RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 10 questions · 24 marks
Slice a cone at different angles and you get a circle, an ellipse, a parabola, or a hyperbola — the four conic sections that describe planetary orbits, satellite dishes, and suspension bridge cables alike. This chapter gives each curve a standard equation centred at the origin and teaches you to read off its key features — centre, radius, focus, directrix, axes — directly from that equation. You will also learn to build the equation from given geometric conditions.
The equation of a circle with centre (2, -3) and radius 5 is:
Answer
(x-2)² + (y+3)² = 25 is correct. Standard form is (x-h)²+(y-k)²=r² with (h,k)=(2,-3) and r=5, so r²=25. This gives (x-2)² + (y-(-3))² = 25, i.e., (x-2)² + (y+3)² = 25.
For the parabola y² = 12x, the coordinates of the focus are:
Answer
The focus is (3, 0). Comparing y²=12x with y²=4ax gives 4a=12, so a=3. For this rightward-opening parabola, the focus is at (a, 0) = (3, 0).
The eccentricity of the ellipse x²/25 + y²/16 = 1 is:
Answer
The eccentricity is 3/5. Here a²=25, b²=16, so a=5, b=4. Then c²=a²-b²=25-16=9, giving c=3. Eccentricity e = c/a = 3/5.
The length of the latus rectum of the parabola x² = 20y is:
Answer
The length is 20. Comparing x²=20y with x²=4ay gives 4a=20, so a=5. Latus rectum length = 4a = 20.
Assertion (A): For the hyperbola x²/9 - y²/16 = 1, the eccentricity is 5/3. Reason (R): For a hyperbola x²/a² - y²/b² = 1, c² = a² + b² and e = c/a.
Answer
Both A and R are true and R is the correct explanation of A. Here a²=9, b²=16, so c²=9+16=25, giving c=5, and a=3. Then e=c/a=5/3, matching Assertion A, which is true. Reason R states the exact formulas c²=a²+b² and e=c/a used to compute this value, so R correctly explains A.
Find the centre and radius of the circle x² + y² - 6x + 8y - 11 = 0.
Answer
Rewrite by completing the square: (x²-6x) + (y²+8y) = 11. (x-3)² - 9 + (y+4)² - 16 = 11. (x-3)² + (y+4)² = 11+9+16 = 36. So the centre is (3, -4) and radius = √36 = 6.
Find the equation of the parabola with vertex at the origin, axis along the y-axis, and passing through the point (6, 3).
Answer
Since the axis is along the y-axis and vertex is at origin, the equation is x² = 4ay (or x²=-4ay if it opens downward). Substituting the point (6,3): 6² = 4a(3), so 36 = 12a, giving a = 3. Since a > 0 and the point has positive y, the parabola opens upward: x² = 12y.
Find the equation of the ellipse with foci at (±4, 0) and vertices at (±5, 0). Also find the length of its latus rectum.
Answer
Step 1 — Since foci and vertices lie on the x-axis, the major axis is along x, so form is x²/a² + y²/b² = 1 with a > b. Step 2 — Vertices at (±5,0) give a = 5, so a² = 25. Step 3 — Foci at (±4,0) give c = 4, so c² = 16. Step 4 — Using c² = a² - b²: 16 = 25 - b², so b² = 9. Step 5 — Equation: x²/25 + y²/9 = 1. Step 6 — Latus rectum = 2b²/a = 2×9/5 = 18/5. So the ellipse is x²/25 + y²/9 = 1 with latus rectum 18/5.
Find the equation of the circle passing through the points (2, 3), (-1, 1) and (3, -1).
Answer
Step 1 — Let the circle be x² + y² + 2gx + 2fy + c = 0. Step 2 — Substitute (2,3): 4+9+4g+6f+c=0 → 4g+6f+c=-13 ... (i) Step 3 — Substitute (-1,1): 1+1-2g+2f+c=0 → -2g+2f+c=-2 ... (ii) Step 4 — Substitute (3,-1): 9+1+6g-2f+c=0 → 6g-2f+c=-10 ... (iii) Step 5 — (i)-(ii): 6g+4f = -11 ... (iv) (iii)-(ii): 8g-4f = -8, i.e., 2g-f=-2 ... (v) Step 6 — From (v): f = 2g+2. Substitute into (iv): 6g+4(2g+2)=-11 → 6g+8g+8=-11 → 14g=-19 → g=-19/14. Then f = 2(-19/14)+2 = -38/14+28/14 = -10/14 = -5/7. From (ii): c = -2+2g-2f = -2+2(-19/14)-2(-5/7) = -2-38/14+10/7 = -2-19/7+10/7 = -2-9/7 = -23/7. Step 7 — Equation: x²+y²+2(-19/14)x+2(-5/7)y-23/7=0, i.e., x²+y² - 19x/7 - 10y/7 - 23/7 = 0, or multiplying by 7: 7x²+7y²-19x-10y-23=0.
Read the following and answer the questions that follow: A satellite dish has a parabolic cross-section. Engineers model it with vertex at the origin and axis along the positive x-axis, given by the equation y² = 8x, with all measurements in centimetres. (a) Find the value of 'a' for this parabola (in the form y² = 4ax). (b) Find the coordinates of the focus, where the receiver should be placed. (c) Find the length of the latus rectum. (d) Find the equation of the directrix.
Answer
(a) Comparing y²=8x with y²=4ax gives 4a=8, so a=2. (b) Focus is at (a,0) = (2,0). (c) Latus rectum = 4a = 8 cm. (d) Directrix is x = -a, i.e., x = -2.
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