Questions: Conic Sections

Q1. Find the centre and radius of the circle x squared plus y squared minus 6x plus 4y minus 12 equal to 0.

Answer: Complete the squares for x and y: x squared minus 6x becomes (x minus 3) squared minus 9, and y squared plus 4y becomes (y plus 2) squared minus 4. The equation becomes (x minus 3) squared plus (y plus 2) squared equal to 25, so the centre is (3, minus 2) and the radius is 5.

Hinglish: Squares complete karo: (x−3)² + (y+2)² = 25, centre (3,−2) aur radius 5.

Q2. Find the focus, directrix and latus rectum of y squared equal to 16x.

Answer: Comparing with y squared equal to 4ax gives 4a equal to 16, so a is 4. The focus is (4, 0), the directrix is x equal to minus 4 and the latus rectum has length 4a equal to 16.

Hinglish: 4a = 16 se a = 4, focus (4,0), directrix x = −4, latus rectum 16.

Q3. Find the eccentricity of the ellipse x squared by 49 plus y squared by 25 equal to 1.

Answer: Here a squared is 49 and b squared is 25, so c squared equals 49 minus 25 equal to 24 and c is 2 root 6. The eccentricity is c by a, which is 2 root 6 by 7.

Hinglish: a² = 49, b² = 25 → c² = 24, c = 2√6, e = 2√6/7.

Q4. Find the foci and eccentricity of the hyperbola x squared by 9 minus y squared by 16 equal to 1.

Answer: Here a is 3 and b is 4, so c squared equals 9 plus 16 equal to 25 and c is 5. The foci are (plus or minus 5, 0) and the eccentricity is 5 by 3, which is greater than 1 as required for a hyperbola.

Hinglish: a = 3, b = 4 → c² = 25, c = 5, foci (±5,0), e = 5/3. Hyperbola me e > 1 hota hai.

Q5. A circle has centre (minus 1, 2) and radius 3. Write its equation.

Answer: Substituting in the standard form gives (x plus 1) squared plus (y minus 2) squared equal to 9. Expanding gives x squared plus y squared plus 2x minus 4y minus 4 equal to 0, which is the general form.

Hinglish: (x+1)² + (y−2)² = 9, expand karo to x² + y² + 2x − 4y − 4 = 0.

Q6. A parabola has focus (2, 0) and directrix x equal to minus 2. Find its equation.

Answer: The vertex is midway between the focus and the directrix, at the origin, and the parabola opens to the right with a equal to 2. The equation is y squared equal to 8x. Checking: a point on the parabola is equidistant from (2, 0) and the line x equal to minus 2, which matches the focus-directrix definition.

Hinglish: Focus (2,0), directrix x = −2 → vertex origin, a = 2, equation y² = 8x.

Q7. Work out the latus rectum length for x squared by 25 plus y squared by 16 equal to 1.

Answer: Here a squared is 25 and b squared is 16, so a is 5 and b is 4. The latus rectum of an ellipse is 2b squared by a, which is 2 times 16 divided by 5, equal to 32 by 5.

Hinglish: a = 5, b = 4, latus rectum = 2b²/a = 32/5.

Q8. Find the asymptotes of the hyperbola x squared by 4 minus y squared by 9 equal to 1.

Answer: Here a is 2 and b is 3, so the asymptotes are y equal to plus or minus b by a times x, that is y equal to plus or minus 3x by 2. The branches approach these lines far from the origin.

Hinglish: a = 2, b = 3 → asymptotes y = ±(b/a)x = ±3x/2.

Q9. A parabola opens upward with its vertex at the origin and passes through (3, 6). Write its equation.

Answer: Opening upward means the equation has the form x squared equal to 4ay. Substituting (3, 6) gives 9 equal to 24a, so a is 3 by 8 and the equation is x squared equal to 3y by 2. The focus is at (0, 3 by 8).

Hinglish: Upward parabola x² = 4ay; (3,6) daalo: 9 = 24a → a = 3/8, equation x² = 3y/2.

Q10. Identify the curve 9x squared plus 16y squared equal to 144 and find its eccentricity.

Answer: Dividing by 144 gives x squared by 16 plus y squared by 9 equal to 1, which is an ellipse with semi-major axis 4 and semi-minor axis 3. Then c squared is 16 minus 9 equal to 7 and the eccentricity is root 7 by 4.

Hinglish: 9x² + 16y² = 144 ko 144 se divide karo: x²/16 + y²/9 = 1, ellipse; c² = 7, e = √7/4.

Before the final answer, run one numerical check: substitute a listed point or focus back into the equation you have written. If the point does not satisfy it, the semi-axes are swapped or a sign has slipped. This check takes seconds and catches most of the slips that cost marks in this chapter.

One habit that pays in conic questions: write the standard form, then write a, b, c and e in a row before touching the question. When a question mixes two curves, list the same four quantities for each curve in separate rows. Writing them out prevents the ellipse and hyperbola relations from getting mixed, which is the single most common cause of a wrong final answer in this chapter. Do the same for the directrix and latus rectum, because those formulas use a and b differently for each curve.

One habit that pays in conic questions: write the standard form, then write a, b, c and e in a row before touching the question. When a question mixes two curves, list the same four quantities for each curve in separate rows. Forgetting to divide by the constant before reading a and b is the second, because coefficients other than 1 hide the semi-axes. Using 2a for the latus rectum of a parabola instead of 4a is the third. Assuming the major axis is always along x is the fourth; compare a and b first. Read the final answer against these four points.

Quiz

The quiz below has five multiple choice questions on the four conics, their eccentricities and latus rectums. Attempt it after the worked questions above..