Conic Sections — Practice

Question 1

Find the equation of a circle with centre (2, -3) and radius 4.

Accepted Answer

(x-2)^2+(y+3)^2=16. Answer: The equation is (x-2)^2+(y+3)^2=16.

Question 2

Find the centre and radius of the circle x^2+y^2-6x+4y-12=0.

Accepted Answer

Completing the square: (x^2-6x+9)+(y^2+4y+4)=12+9+4, so (x-3)^2+(y+2)^2=25. Answer: Centre (3,-2), radius 5.

Question 3

Find the focus, directrix, and length of the latus rectum of y^2=12x.

Accepted Answer

Comparing with y^2=4ax: 4a=12, so a=3. Answer: Focus (3,0), directrix x=-3, latus rectum 4a=12.

Question 4

Find the equation of the parabola with vertex at the origin, axis along the y-axis, and passing through (4, -2).

Accepted Answer

Since it passes through a point with negative y, the parabola opens downward: x^2=-4ay. Substituting (4,-2): 16=-4a(-2)=8a, so a=2. Answer: The equation is x^2=-8y.

Question 5

Find the equation of the ellipse with foci (+-3, 0) and semi-major axis a=5.

Accepted Answer

c=3, a=5, so b^2=a^2-c^2=25-9=16. Answer: The equation is x^2/25+y^2/16=1.

Question 6

Find the coordinates of the foci, the eccentricity, and the latus rectum of the ellipse x^2/16+y^2/25=1.

Accepted Answer

Since 25>16, the major axis is along the y-axis with a=5, b=4. c=sqrt(25-16)=3. Answer: Foci (0,+-3), eccentricity 3/5, latus rectum 2(16)/5=32/5.

Question 7

For the hyperbola x^2/9-y^2/16=1, find the eccentricity, foci, and latus rectum.

Accepted Answer

a^2=9, b^2=16, so c^2=a^2+b^2=25, c=5. Answer: Eccentricity e=c/a=5/3, foci (+-5,0), latus rectum 2(16)/3=32/3.

Question 8

Find the equation of the hyperbola with vertices (0, +-3) and foci (0, +-5).

Accepted Answer

a=3, c=5, so b^2=c^2-a^2=25-9=16. Since vertices are on the y-axis, the transverse axis is vertical. Answer: The equation is y^2/9-x^2/16=1.

Question 9

Find the equation of the circle passing through (0,0) and making intercepts 4 and 6 on the coordinate axes.

Accepted Answer

A circle through the origin with x-intercept 4 and y-intercept 6 also passes through (4,0) and (0,6). Using the general form x^2+y^2+2gx+2fy+c=0: since it passes through (0,0), c=0. Through (4,0): 16+8g=0, so g=-2. Through (0,6): 36+12f=0, so f=-3. Answer: The equation is x^2+y^2-4x-6y=0.

Question 10

Find the equation of the ellipse with major axis along the x-axis, passing through the points (4,3) and (-1,4).

Accepted Answer

Using x^2/a^2+y^2/b^2=1: from (4,3), 16/a^2+9/b^2=1; from (-1,4), 1/a^2+16/b^2=1. Solving this pair simultaneously (let u=1/a^2, v=1/b^2): 16u+9v=1 and u+16v=1. From the second, u=1-16v; substituting: 16(1-16v)+9v=1, so 16-256v+9v=1, giving -247v=-15, v=15/247, and then u=7/247. Answer: The equation is 7x^2+15y^2=247.

Practice — Conic Sections

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