If a circle of radius 5 is tangent to the x-axis at the origin, its centre must lie vertically above or below the tangent point: (0,5) or (0,-5). Hence (x-0)^2+(y-5)^2=25 and (x-0)^2+(y+5)^2=25. Expanding gives x^2+y^2-10y=0 and x^2+y^2+10y=0.
Two circles: x^2+(y-5)^2=25 and x^2+(y+5)^2=25. In general form: x^2+y^2-10y=0 and x^2+y^2+10y=0.
Radius squared r^2 = (3-2)^2+(6-1)^2 =1+25=26. Therefore (x-2)^2+(y-1)^2=26.
(x-2)^2+(y-1)^2=26.
If a circle touches both axes its centre is (a,−a) or equivalents with equal magnitude coordinates. Let centre = (a,−a) and radius = |a|. Plugging (-4,2) into (x-a)^2+(y+a)^2=a^2 gives (a+4)^2+(a+2)^2=a^2 ⇒ a^2+12a+20=0 ⇒ a=-10 or a=-2. Thus centres (-10,10) with r=10 and (-2,2) with r=2. Equations as above.
Two circles: (x+10)^2+(y-10)^2=100 and (x+2)^2+(y-2)^2=4. In general form: x^2+y^2+20x-20y=0 and x^2+y^2+4x-4y=0.
Intersection: solve 3x-2y-1=0 and 4x+2y-7=0. Adding gives 7x-8=0 ⇒ x=8/7; then y=(3x-1)/2=17/14. Radius^2 = (2-8/7)^2+(3-17/14)^2 = (6/7)^2+(25/14)^2 =36/49+625/196=769/196. Circle: (x-2)^2+(y-3)^2=769/196. (If the second line was different, recompute intersection accordingly.)
(x-2)^2+(y-3)^2 = 769/196.
Centre is midpoint: ((3+2)/2,(4+7)/2)=(5/2,11/2). Radius^2 = (3-5/2)^2+(4-11/2)^2 = (1/2)^2+(-3/2)^2 =1/4+9/4=10/4=5/2. Hence equation as above.
(x-5/2)^2+(y-11/2)^2=5/2.
Let the general equation of the circle be
x² + y² + 2gx + 2fy + c = 0
It passes through the points (1, 0), (-1, 0) and (0,1)
1 + 0 + 2g + c = 0
2g + c = -1 ………(1)
1 + 0 – 2g + c = 0
-2g + c = -1 …………(2)
0 + 1 + 0 + 2f + c = 0
2f + c = -1
(1) + (2) ⇒ 2c = -2
c = -1
substitute in eqn (1)
2g – 1 = -1
2g = 0
g = 0
substitute in eqn (3)
2f – 1 = -1
2f = -1 + 1
2f = 0
f = 0
Therefore, the required equation of the circle
x² + y² – 1 = 0
f = 0 Therefore, the required equation of the circle x² + y² – 1 = 0
General circle x^2+y^2+2gx+2fy+c=0. Plug (1,0): 1+2g+c=0. Plug (-1,0):1-2g+c=0. Subtract ⇒ g=0. Then c=-1. Plug (0,1):1+2f-1=0 ⇒ f=0. So x^2+y^2-1=0.
x^2+y^2-1=0.
Area 9π ⇒ r^2=9 ⇒ r=3. If two diameters lie along the given lines, the centre is their intersection. Solve x+y=5, x−y=1 ⇒ x=3,y=2. So circle (x-3)^2+(y-2)^2=9.
(x-3)^2+(y-2)^2=9.
Distance from centre (0,0) to line x - y + c =0 is |c|/√2. For tangency this equals radius 4 ⇒ |c|/√2 =4 ⇒ |c|=4√2 ⇒ c=±4√2.
c = ±4√2.
Substitute (2,2) into the left-hand side: 4+4+4-12-12-8 = -20 ≠ 0. Hence (2,2) is not on the circle; therefore no tangent or normal at that point to the given circle.
Point (2,2) is not on the given circle, so no tangent/normal there.
For f(x,y)=x^2+y^2+xy-2x-5y-5, a point is inside iff f<0, on iff f=0, outside iff f>0. Compute: (-2,1): 4+1-2+4-5-5 = -3 (<0) ⇒ inside. (0,0): -5 (<0) ⇒ inside. (-4,3):16+9-12+8-15-5 =1 (>0) ⇒ outside.
(-2,1): inside; (0,0): inside; (-4,3): outside.
(i) x² + (y + 2)² = 0
(ii) x² + y² + 6x – 4y + 4 = 0
(iii) x² + y² – x + 2y – 3 = 0
(iv) 2x² + 2y² – 6x + 4y + 2 = 0v
(i) x 2 + (y + 2) 2 = 0
(i.e) x 2 + y 2 + 4y + 4 = 0
Comparing this equation with the general form x 2 + y 2 + 2gx + 2fy + c = 0
we get 2g = 0 ⇒ g = 0
2f = 4 ⇒ f= 2 and c = 4
Now centre = (-g, -f) = (0, -2)
Radius = r = $\sqrt{g^{2}+f^{2}-c}=\sqrt{0+4-4}$
∴ Centre = (0, -2) and radius = 0
(ii) x 2 + y 2 + 6x – 4y + 4 = 0
Comparing with the general form we get
2g = 6, 2f = -4
⇒ g = 3, /= -2 and c = 4
Centre = (-g, -f) = (-3, 2)
Radius = $\sqrt{g^{2}+f^{2}-c}=\sqrt{9+4-4}$= 3
∴ Centre = (-3, 2) and radius = 3
(iii) x² + y² – x + 2y – 3 = 0
2g = -1; 2f = 2; c = -3
g = $\frac{-1}{2}$ f = 1
Centre (-g, -f) = ($\frac{1}{2}$, -1)
Radius = $\sqrt{g^2+f^2-c}$ = $\sqrt{\frac{1}{4}+1+3}$
$\sqrt{\frac{1+4+12}{4}}$
$\sqrt{\frac{17}{4}}$ = $\frac{\sqrt{17}}{2}$
(iv) 2x 2 + 2y 2 – 6x + 4y + 2 = 0
(÷ by 2) ⇒ x 2 + y 2 – 3x + 2y + 1 =0
Comparing this equation with the general form of the circle we get
2g = -3, 2f= 2
g = $-\frac{3}{2}$, g= 1 and c = 1
So centre = (-g, -f) = ($\frac{3}{2}$, -1)
and radius = $\sqrt{g^{2}+f^{2}-c}=\sqrt{\frac{9}{4}+1-1}=\frac{3}{2}$
∴ Centre = ($\frac{3}{2}$, -1) and radius = $\frac{3}{2}$
So centre = (-g, -f) = ($\frac{3}{2}$, -1) and radius = $\sqrt{g^{2}+f^{2}-c}=\sqrt{\frac{9}{4}+1-1}=\frac{3}{2}$ ∴ Centre = ($\frac{3}{2}$, -1) and radius = $\frac{3}{2}$
3x² + (3 – p) xy + qy² – 2px = 8pq represent a circle means,
Co-efficient of x² = co-efficient of y²
3 = q ⇒ q = 3
Co-efficient of xy = 0
3 – p = 0 ⇒ p = 3
3x² + 3y² – 6x = 8 (3)(3)
3x² + 3y² – 6x – 72 = 0
(÷3) x² + y² – 2x – 24 = 0
2g = -2; 2f = 0; c = -24
g = -1 f = 0
Centre (-g, -f) = (1, 0)
Radius = $\sqrt{g^2+f^2-c}$ = $\sqrt{1+0+24}$
= $\sqrt{25}$ = 5
Centre (-g, -f) = (1, 0) Radius = $\sqrt{g^2+f^2-c}$ = $\sqrt{1+0+24}$ = $\sqrt{25}$ = 5
(i) Equidistance from (4,0) and line x=-4: (x-4)^2+y^2=(x+4)^2 ⇒ y^2=16x. (ii) Parabola symmetric about y-axis: x^2=4ay. Plug (-2,3):4=12a ⇒ a=1/3 ⇒ x^2=(4/3)y. (iii) Focus right of vertex by p=3 ⇒ (y-2)^2=4p(x-1)=12(x-1). (iv) Midpoint of latus rectum is focus (4,0). Latus rectum half-length 8 ⇒ 2p=8 ⇒ p=4, focus at (h+p,k)=(4,0) ⇒ h=0,k=0. Equation y^2=4p(x-h)=16x.
(i) y^2=16x. (ii) x^2=(4/3)y. (iii) (y-2)^2=12(x-1). (iv) y^2=16x.
(i) foci (± 3, 0), e = $\frac {1}{2}$
(ii) foci (0, ±4) and end points of major axis are (0, ±5).
(iii) length of latus rectum 8, eccentricity = $\frac {3}{5}$ and major axis on x-axis.
(iv) length of latus rectum 4, distance between foci and major axis as y axis.v
(i) foci (± 3, 0), e = $\frac {1}{2}$
foci (± c, 0) = (± 3, 0)
e = $\frac {1}{2}$
c = ae = 3
a($\frac {1}{2}$) = 3
a = 6 ⇒ a² = 36
b² = a² – c²
b² = 36 – 9 = 27
b² = 27
Equation of the ellipse be $\frac {x^2}{a^2}$ + $\frac {y^2}{b^2}$ = 1
$\frac {x^2}{36}$ + $\frac {y^2}{27}$ = 1
(ii) foci (0, ±4) and end points of major axis are (0, ±5)
foci (0, ±c) = (0, +4)
vertex (0, ±a) = (0, ±5)
∴ c = 4, a = 5
ae = 4
5e = 4
e = $\frac {4}{5}$
b² = a² – c²
= 25 – 16
b² = 9
Equation of the ellipse be $\frac {x^2}{b^2}$ + $\frac {y^2}{a^2}$ = 1
$\frac {x^2}{9}$ + $\frac {y^2}{25}$ = 1
(iii) length of latus rectum 8, eccentricity = $\frac {3}{5}$ and major axis on x-axis.
e = $\frac {3}{5}$
Latus rectum $\frac {2b^2}{a}$ = 8
(iv) length of latus rectum 4, distance between foci 4√2 and major axis as y-axis
Given $\frac{2 b^{2}}{a}$ = 4 and 2ae = $4 \sqrt{2}$
Now $\frac{2 b^{2}}{a}$ = 4 2b 2 = 4a
⇒ b 2 = 2a
2ae = $4 \sqrt{2}$ ae = $2 \sqrt{2}$
So a 2 e 2 = 4(2) = 8
We know b 2 = a 2 (1 – e 2 ) = a 2 – a 2 e 2
⇒ 2a = a 2 – 8 ⇒ a 2 – 2a -8 = 0
⇒ (a – 4) (a +2) = 0 ⇒a = 4 or -2
As a cannot be negative
a = 4 So a 2 = 16 and b 2 = 2(4) = 8
Also major axis is along j-axis
So equation of ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{16}$ = 1
a = 4 So a 2 = 16 and b 2 = 2(4) = 8 Also major axis is along j-axis So equation of ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{16}$ = 1
(i) foci (± 2, 0), eccentricity = $\frac {3}{2}$
(ii) centre (2, 1) one of the foci (8, 1) and corresponding directrix x = 4.
(iii) passing through (5, -2) and length of the transverse axis along x axis and length 8 units.v
(i) foci (± 2, 0), eccentricity = $\frac {3}{2}$
(ii) centre (2, 1) one of the foci (8, 1) and corresponding directrix x = 4.
Distance CS = ae = 6 …….. (1)
Directrix $\frac {a}{e}$ = 4 ……… (2)
(1) × (2) ⇒ ae × $\frac {a}{e}$ = 24
a² = 24
∴ c = ae = 6
b² = c² – a²
= 36 – 24 = 12
The transverse axis is parallel to x-axis
∴ $\frac {(x-h)^2}{a^2}$ – $\frac {(y – k)^2}{b^2}$ = 1 (h, k) = (2, 1)
$\frac {(x-2)^2}{24}$ – $\frac {(y – 1)^2}{12}$ = 1
(iii) passing through (5,-2) and length of the transverse axis along x axis and of length 8 units.
Transverse axis along x-axis
$\frac {x^2}{a^2}$ – $\frac {y^2}{b^2}$ = 1
Length of transverse axis 2a = 8
⇒ a = 4
$\frac {x^2}{a^2}$ – $\frac {y^2}{b^2}$ = 1 Length of transverse axis 2a = 8 ⇒ a = 4
(i) foci (± 2, 0), eccentricity = $\frac {3}{2}$
(ii) centre (2, 1) one of the foci (8, 1) and corresponding directrix x = 4.
(iii) passing through (5, -2) and length of the transverse axis along x axis and length 8 units.v
(i) foci (± 2, 0), eccentricity = $\frac {3}{2}$
(ii) centre (2, 1) one of the foci (8, 1) and corresponding directrix x = 4.
Distance CS = ae = 6 …….. (1)
Directrix $\frac {a}{e}$ = 4 ……… (2)
(1) × (2) ⇒ ae × $\frac {a}{e}$ = 24
a² = 24
∴ c = ae = 6
b² = c² – a²
= 36 – 24 = 12
The transverse axis is parallel to x-axis
∴ $\frac {(x-h)^2}{a^2}$ – $\frac {(y – k)^2}{b^2}$ = 1 (h, k) = (2, 1)
$\frac {(x-2)^2}{24}$ – $\frac {(y – 1)^2}{12}$ = 1
(iii) passing through (5,-2) and length of the transverse axis along x axis and of length 8 units.
Transverse axis along x-axis
$\frac {x^2}{a^2}$ – $\frac {y^2}{b^2}$ = 1
Length of transverse axis 2a = 8
⇒ a = 4
$\frac {x^2}{a^2}$ – $\frac {y^2}{b^2}$ = 1 Length of transverse axis 2a = 8 ⇒ a = 4
(i) y² = 16x
(ii) x² = 24y
(iii) y² = -8x
(iv) x² – 2x + 8y + 17 = 0
(v) y² – 4y – 8x + 12 = 0v
(i) y² = 16x
4a = 16
a = 4
(a) Vertex V (0, 0)
(b) Focus S(a, 0) = S(4, 0)
(c) Equation of the directrix x = – a
x = -4 ⇒ x + 4 = 0
(d) Length of the latus rectum = 4a = 4(4)
= 16
(ii) x² = 24y
(a) Vertex V (0, 0),
(b) Focus S (0, a) = S(0, 6)
(c) Equation of the directrix y = -a = -6
⇒ y + 6 = 0
(d) Length of the latus rectum = 4a
= 4 (6) = 24
(iii) y² = -8x
4a = 8,
a = 2
(a) Vertex V(0, 0) = ( 0, 0)
(b) Focus S(-a, 0) = (-2, 0)
(c) Equation of the directrix x = a = 2
x – 2 = 0
(d) Length of the latus rectum 4a = 8
(iv) x² – 2x + 8y + 17 = 0
x² – 2x = -8y – 17
(x – 1)² = -8y – 17 + 1
(x – 1)² = -8y – 16
(x – 1)² = -8(y + 2)
It is form of (x – h)² = -4a(y – k)
4a = 8 ⇒ a = 2
(a) Vertex be (h, k) = (1, -2)
(b) Foeus = (0 +h, -a + k) = (0 + 1, -2 – 2) = (1, -4)
(c) Equation of the directrix is y + k + a = 0
y – 2 + 2= 0
y = 0
(d) Length of latus rectum is 4a = 4 × 2 = 8 units,
(v) y² – 4y – 8x + 12 = 0
y² – 4y = 8x – 12
(y – 2)² = 8x – 12 + 4
= 8x – 8
= 8 (x – 1)
(y – 2)² = 8 (x – 1)
It is form of (y – k)² = Aa(x – h)
4a = 8 ⇒ a = 2
(a) Vertex (h, k) = (1, 2)
(b) Focus = (a+h, 0 + k) = (2 + 1, 0 + 2) = (3, 2)
(c) Equation of the directrix x = -a + h
= -2 + 1
= -1
x + 1 = 0
(d) Length of latus rectum is
4a = 4 × 2 = 8 units.
x + 1 = 0 (d) Length of latus rectum is 4a = 4 × 2 = 8 units.
(i) $\frac {x^2}{25}$ + $\frac {y^2}{9}$ = 1
(ii) $\frac {x^2}{3}$ + $\frac {y^2}{10}$ = 1
(iii) $\frac {x^2}{25}$ – $\frac {y^2}{144}$ = 1
(iv) $\frac {y^2}{16}$ – $\frac {x^2}{9}$ = 1v
(i) $\frac{x^{2}}{25}+\frac{y^{2}}{9}$ = 1
It is of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ = 1, which is an ellipse
Here a 2 = 25, b 2 = 9
a = 5, b = 3
e 2 = $\frac{a^{2}-b^{2}}{a^{2}}=\frac{25-9}{25}=\frac{16}{25}$ ⇒ e = $\frac{4}{5}$
Now e = $\frac{4}{5}$ and a = 5 ⇒ ae = 4 and $\frac{a}{e}=\frac{5}{4 / 5}=\frac{25}{4}$
Here the major axis is along x axis
∴ Centre = (0, 0)
Foci = (± ae, 0) = (± 4, 0)
Vertices = (± a, 0) = (±5, 0)
Equation of directrix x = ± $\frac{a}{e}$ (ie.,) x = ± $\frac{25}{4}$
(ii) $\frac {x^2}{3}$ + $\frac {y^2}{10}$ = 1
It is an ellipse. The major axis is along y axis
a² = 10 b² = 3
a = $\sqrt {10}$ b = √3
c² = a² – b²
= 10 – 3 = 7
c = √7
ae = √7.
$\sqrt {10}$ = √7
e = $\sqrt {\frac{7}{10}}$
(a) Centre (0, 0)
(b) Vertex (0, ±a) = (0, ±$\sqrt {10}$)
(c) Foci (0, ±c) – (0, ±√7)
(d) Equation of the directrix a
y = ±$\frac{a}{e}$
= ±$\frac{\sqrt {10}}{√7}$.$\sqrt {10}$ = ±$\frac {10}{√7}$
y = ±$\frac {10}{√7}$
(iii) $\frac {x^2}{25}$ – $\frac {y^2}{144}$ = 1
It is Hyperbola. The transverse axis the x axis.
a² = 25; b² = 144
a = 5; b = 12
c² = a² + b²
= 25 + 144 = 169
c = 13
ae = 13
5e = 13
e = $\frac {13}{5}$
(a) Centre (0, 0)
(b) Vertex (± a, 0) = (± 5, 0)
(c) Foci (± c, 0) = (± 13, 0)
(d) Equation of the directrix
x = ±$\frac {a}{e}$ = ±$\frac {5}{\frac{13}{5}}$ = ±$\frac {25}{13}$
x = ±$\frac {25}{13}$
(iv) $\frac {y^2}{16}$ – $\frac {x^2}{9}$ = 1
It is Hyperbola. The transverse axis the y axis.
a² = 16; b² = 9
a = 4; b = 3
c² = a² + b²
= 16 + 6 = 25
c = 5
ae = 5
4e = 5
e = $\frac {5}{4}$
(a) Centre (0, 0)
(b) Vertex (0, ±a) = (0, ±4)
(c) Foci (0, ±ae) = (0, ±5)
(d) Equation of the directrix
y = ±$\frac {a}{e}$ = ±$\frac {4}{\frac{5}{4}}$ = ±$\frac {16}{5}$
y = ±$\frac {16}{5}$
(d) Equation of the directrix y = ±$\frac {a}{e}$ = ±$\frac {4}{\frac{5}{4}}$ = ±$\frac {16}{5}$ y = ±$\frac {16}{5}$
For hyperbola x^2/a^2 - y^2/b^2 =1, focus at (c,0) with c^2 = a^2 + b^2. Equation of a vertical line through focus x=c. Intersect with hyperbola: c^2/a^2 - y^2/b^2 =1 ⇒ y^2/b^2 = c^2/a^2 -1 = (c^2 - a^2)/a^2 = b^2/a^2. Hence y^2 = b^4/a^2 ⇒ y = ± b^2/a. Distance between these two points (the latus rectum) is 2b^2/a.
Length of latus rectum = 2b^2/a.
Compute \[d_1-d_2=\sqrt{(x-c)^2+y^2}-\sqrt{(x+c)^2+y^2}.\] Square and simplify twice (or use the defining equation) to obtain \(|d_1-d_2|=2a\). Thus the absolute difference of focal distances equals the transverse axis length \(2a\).
For the standard hyperbola \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) with foci \((\pm c,0),\;c^2=a^2+b^2\), any point \(P(x,y)\) has focal distances \(d_1=\sqrt{(x-c)^2+y^2}\) and \(d_2=\sqrt{(x+c)^2+y^2}\). For a point on the hyperbola one gets
Answer:
(i) Ellipse, centre $(3,4)$; vertices $(3,21),(3,-13)$; foci $(3,12),(3,-4)$; directrices $y=4\pm\frac{289}{8}$.
(ii) Ellipse, centre $(-1,2)$; vertices $(9,2),(-11,2)$; foci $(5,2),(-7,2)$; directrices $x=-1\pm\frac{50}{3}$.
(iii) Hyperbola, centre $(-3,4)$; vertices $(12,4),(-18,4)$; foci $(14,4),(-20,4)$; directrices $x=-3\pm\frac{225}{17}$.
(iv) Hyperbola, centre $(-1,2)$; vertices $(-1,7),(-1,-3)$; foci $(-1,2+\sqrt{41}),(-1,2-\sqrt{41})$; directrices $y=2\pm\frac{25}{\sqrt{41}}$.
(v) $18x^2+12y^2-144x+48y+120=0$ becomes $\frac{(x-4)^2}{12}+\frac{(y+2)^2}{18}=1$. It is an ellipse, centre $(4,-2)$; vertices $(4,-2\pm3\sqrt2)$; foci $(4,-2\pm\sqrt6)$; directrices $y=-2\pm3\sqrt6$.
(vi) $9x^2-y^2-36x-6y+18=0$ becomes $(x-2)^2-\frac{(y+3)^2}{9}=1$. It is a hyperbola, centre $(2,-3)$; vertices $(3,-3),(1,-3)$; foci $(2\pm\sqrt{10},-3)$; directrices $x=2\pm\frac{1}{\sqrt{10}}$.
(5) 11x² – 25y² – 44x + 50y – 256 = 0
(6) y² + 4x + 3y + 4 = 0
(1) 2x² – y² = 7 (2) 3x² + 3y² – 4x + 3y + 10 = 0 (3) 3x² + 2y² = 14 (4) x² + y² + x – y = 0v
(1) 2x² – y² = 7
Comparing this equation with the general equation of the conic
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
We get A = 2, C = – 1
Elere A ≠ C also A and C are of opposite signs.
So the conic is a hyperbola.
(2) 3x² + 3y² – 4x + 3y + 10 = 0
A = 3, B = 0, C = 3, D = -4, E = 3, F = 10
A = C and B = 0 (No xy term)
∴ It is a circle.
(3) 3x² + 2y² = 14
A = 3, B = 0, C = 2, F = -14
A ≠ C and A & C are the same signs.
∴ It is an ellipse.
(4) x² + y² + x – y = 0
Comparing this equation with the general equation of the conic
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
We get A = C and B = 0
So the given conic is a circle.
(5) 11x² – 25y² – 44x + 50y – 256 = 0
A =11, B = 0, C = -25, D = -44, E = 50, F = -256
A ≠ C and A & C are the opposite signs.
∴ It is a hyperbola.
(6) y² + 4x + 3y + 4 = 0
Comparing this equation with the general equation of the conic
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
We get A = 0 and B = 0
So the conic is a parabola.
2x² – y² = 7
(5) 11x² – 25y² – 44x + 50y – 256 = 0
(6) y² + 4x + 3y + 4 = 0
(1) 2x² – y² = 7 (2) 3x² + 3y² – 4x + 3y + 10 = 0 (3) 3x² + 2y² = 14 (4) x² + y² + x – y = 0v
(1) 2x² – y² = 7
Comparing this equation with the general equation of the conic
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
We get A = 2, C = – 1
Elere A ≠ C also A and C are of opposite signs.
So the conic is a hyperbola.
(2) 3x² + 3y² – 4x + 3y + 10 = 0
A = 3, B = 0, C = 3, D = -4, E = 3, F = 10
A = C and B = 0 (No xy term)
∴ It is a circle.
(3) 3x² + 2y² = 14
A = 3, B = 0, C = 2, F = -14
A ≠ C and A & C are the same signs.
∴ It is an ellipse.
(4) x² + y² + x – y = 0
Comparing this equation with the general equation of the conic
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
We get A = C and B = 0
So the given conic is a circle.
(5) 11x² – 25y² – 44x + 50y – 256 = 0
A =11, B = 0, C = -25, D = -44, E = 50, F = -256
A ≠ C and A & C are the opposite signs.
∴ It is a hyperbola.
(6) y² + 4x + 3y + 4 = 0
Comparing this equation with the general equation of the conic
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
We get A = 0 and B = 0
So the conic is a parabola.
2x² – y² = 7
Write ellipse in standard form: x^2/7 + y^2/2 = 1 (so a^2=7, b^2=2). A line with slope m: y=mx+c is tangent iff c^2 = a^2 m^2 + b^2 = 7m^2+2. Since line passes (5,2), c=2-5m. Hence (2-5m)^2 = 7m^2+2 => 9m^2-10m+1=0 => (9m-1)(m-1)=0 so m=1 or m=1/9. For m=1, c= -3 gives y = x-3. For m=1/9, c=13/9 gives 9y = x+13 or x -9y +13=0.
The two tangents are y = x - 3 and x - 9y + 13 = 0.
Given slope m = 10/3. For hyperbola x^2/a^2 - y^2/b^2 =1 (a^2=16,b^2=64) the tangent with slope m has form y = mx ± \sqrt{a^2 m^2 - b^2}. Here a^2 m^2 - b^2 =16*(100/9)-64 =1024/9, so \sqrt{ } =32/3. Thus lines: y = (10/3)x ± 32/3, equivalently 10x-3y \pm 32 =0.
The required tangents are 10x - 3y + 32 = 0 and 10x - 3y - 32 = 0.
y 2 = 8x
Comparing this equation with y 2 = 4ax
we get 4a = 8 ⇒ a = 2
Now, the parametric form for y 2 = 4ax is x = at 2, y = 2at
Here a = 2 and t = 2
⇒ x = 2(2) 2 = 8 and y = 2(2) (2) = 8
So the point is (8, 8)
Now eqution of tangent to y 2 = 4 ax at (x 1, y 1 ) is yy 1 = 2a(x + x 1 )
Here (x 1, y 1 ) = (8, 8) and a = 2
So equation of tangent is y(8) = 2(2) (x + 8)
(ie.,) 8y = 4 (x + 8)
(÷ by 4) ⇒ 2y = x + 8 ⇒ x – 2y + 8 = 0
Aliter
The equation of tangent to the parabola y 2 = 4ax at ‘t’ is
yt = x + at 2
Here t = 2 and a = 2
So equation of tangent is
(i.e.,) y(2) = x + 2(2) 2
2y = x + 8 ⇒ x – 2y + 8 = 0
So equation of tangent is (i.e.,) y(2) = x + 2(2) 2 2y = x + 8 ⇒ x – 2y + 8 = 0
The ellipse is x^2 + 3y^2 = 12, or x^2/12 + y^2/4 = 1. On the line x - y + 4 = 0, we have x = y - 4. Substitute in the ellipse:
(y - 4)^2 + 3y^2 = 12
y^2 - 8y + 16 + 3y^2 = 12
4y^2 - 8y + 4 = 0
(y - 1)^2 = 0.
The repeated root shows that the line meets the ellipse at exactly one point, so it is a tangent. When y = 1, x = -3.
Parametric check: for x^2/12 + y^2/4 = 1, take a = 2sqrt(3), b = 2. The point (-3, 1) lies on the ellipse, and the tangent formula xx1/a^2 + yy1/b^2 = 1 gives -x/4 + y/4 = 1, i.e. x - y + 4 = 0.
The line is a tangent, and the point of contact is (-3, 1).
Equation of the parabola
y² = 16x
4 a = 16
a = 4
Tangent is perpendicular to the line
2x + 2y + 3 = 0 is 2x – 2y + k = 0
2x – 2y + k = 0
2y = 2x + k
y = x + $\frac {k}{2}$
m = 1 c = $\frac {k}{2}$
Condition that the line y = mx + c to be tangent to the parabola is
c = $\frac {a}{m}$
$\frac {k}{2}$ = $\frac {4}{1}$
k = 8
Equation of the tangent
2x – 2y + 8 = 0
÷ by 2 ⇒ x – y + 4 = 0
Equation of the tangent 2x – 2y + 8 = 0 ÷ by 2 ⇒ x – y + 4 = 0
For y^2=4ax, here 4a=8 so a=2. Parametric point for parameter t is (at^2,2at) = (2t^2,4t). For t=2 the point is (8,8). Tangent: ty = x + at^2 => 2y = x + 8 => x -2y +8 =0.
x - 2y + 8 = 0.
(Hint: use parametric form)v
(i) Equation of the tangent to hyperbola be
⇒ 4x – 3y = 6
⇒ 4x – 3y – 6 = 0
(ii) Equation of the normal to hyperbola be
⇒ 3x + 4y – 42 = 0
⇒ 4x – 3y – 6 = 0 (ii) Equation of the normal to hyperbola be ⇒ 3x + 4y – 42 = 0
For parameter t the tangent to y^2 = 4ax is: t y = x + a t^2.
For t = t1 and t = t2 we have the two equations:
t1 y = x + a t1^2
t2 y = x + a t2^2
Subtracting gives (t1 − t2) y = a(t1^2 − t2^2) = a(t1 − t2)(t1 + t2), so y = a(t1 + t2).
Substitute y into one tangent equation, say t1 y = x + a t1^2: x = t1·a(t1 + t2) − a t1^2 = a t1 t2.
Hence the intersection point is (a t1 t2, a(t1 + t2)).
The normal at $t_1$ is $y+t_1x=2at_1+at_1^3$. Since it passes through $(at_2^2,2at_2)$: $2at_2+t_1\,at_2^2=2at_1+at_1^3$. Dividing by $a$: $t_1t_2^2+2t_2-2t_1-t_1^3=0\Rightarrow t_1(t_2^2-t_1^2)+2(t_2-t_1)=0\Rightarrow(t_2-t_1)\big[t_1(t_2+t_1)+2\big]=0$. As $t_2\ne t_1$, $t_1(t_1+t_2)+2=0$, giving $t_2=-t_1-\dfrac{2}{t_1}$.
$t_2=-t_1-\dfrac{2}{t_1}$ (proved).
Place vertex at (0,10). The parabola passes through (15,0). Equation y = kx^2 +10 gives 0 = k(225)+10 ⇒ k = -10/225 = -2/45. At x=6, y = -2/45(36)+10 = -72/45 +10 = 8.4 m.
8.4 m.
Ellipse centered at origin with semi‑vertical b=5. At horizontal distance x=8 (edge of 16 m road) require y=4: 4 = 5 sqrt(1 - 64/a^2). Squaring: 16/25 = 1 - 64/a^2 ⇒ 64/a^2 = 9/25 ⇒ a^2 = 1600/9 ⇒ a = 40/3. Opening width = 2a = 80/3 ≈ 26.667 m.
The opening must be 80/3 m ≈ 26.667 m wide.
Vertex is at (0.5,4). Parabola: y = k(x-0.5)^2 + 4. Since it passes through (0,0): 0 = k(0.25)+4 ⇒ k = -16. Thus y = -16(x-0.5)^2 +4. At x=0.75, (x-0.5)=0.25 ⇒ y = -16(0.0625)+4 = -1 +4 = 3 m.
3 m.
(a) With vertex at origin and axis along x, focus at (p,0) with p=1.2 ⇒ standard form x = (1/(4p))y^2 = 1/4.8 y^2 = (5/24) y^2. (b) Rim has y = ±2.5 m, so depth x = (1/4.8)*(2.5)^2 = 6.25/4.8 ≈ 1.30208 m.
v
Equation of hyperbola is $\frac {x^2}{30^2}$ – $\frac {y^2}{44^2}$ = 1
Given OC = $\frac {1}{2}$ OD and CD = 150
∴ OC = 50 m and OD = 100 m
Let the Radius of top of the tower be x 1 and bottom of the tower be x 2.
∴ Points A(x 1, 50) and B(x 2, 100)
Hyperbola passes through A(x 1, 50)
∴ Radius of the top = 45.41 m.
Diameter of the top = 90.82 m
Also
The hyperbola again passes through B(x 2, 100)
∴ Radius of the base = 74.48 m.
Diameter of the base = 148.96 m
∴ Diameter of the top and base of the tower are 90.82 m and 148.96 m.
∴ Radius of the base = 74.48 m. Diameter of the base = 148.96 m ∴ Diameter of the top and base of the tower are 90.82 m and 148.96 m.
Let the rod endpoints be (x,0) and (0,y) with x^2+y^2=(1.2)^2=1.44. The point P is 0.3 m from the x-axis end along the rod of length 1.2, so its parametric position is P = (0.75x,0.25y). Let X=0.75x, Y=0.25y ⇒ x=(4/3)X, y=4Y. Substitute into x^2+y^2=1.44: (16/9)X^2 + 16Y^2 = 1.44. Divide both sides by 1.44 to get standard ellipse: X^2/(0.81) + Y^2/(0.09) = 1, so a^2 = 0.81, b^2 = 0.09. Hence e = sqrt(1 - b^2/a^2) = sqrt(1 - 0.09/0.81) = sqrt(8/9) = 2√2/3.
e = (2√2)/3.
With vertex at (0,0) the parabola is y = kx^2. Using (3,−2.5): −2.5 = k·9 ⇒ k = −2.5/9 = −5/18. At ground y = −7.5: −7.5 = (−5/18)x^2 ⇒ x^2 = 27 ⇒ x = 3√3. Hence the water strikes at 3√3 m beyond the vertical.
3\sqrt{3} m
Parabola with zeros at x=0 and x=12 and vertex at x=6: y = kx(12−x). Max at x=6 gives 36k = 4 ⇒ k = 1/9. So y = (1/9)x(12−x) = (12/9)x − (1/9)x^2. Compare with y = x tanθ − (g/(2u^2 cos^2θ))x^2 ⇒ tanθ = 12/9 = 4/3. Hence θ = arctan(4/3).
\theta = \arctan\frac{4}{3}
Place A at (−5,0), B at (5,0). Condition: difference of distances to foci is 6, i.e. |PA − PB| = 6 ⇒ 2a = 6 ⇒ a = 3. For hyperbola c = 5, so b^2 = c^2 − a^2 = 25 − 9 = 16. Equation: x^2/a^2 − y^2/b^2 = 1 ⇒ x^2/9 − y^2/16 = 1.
Hyperbola: \displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1
General member: x^2+y^2+(5+4λ)x+(-6+3λ)y+(9−19λ)=0. For tangency to y‑axis, if general form is x^2+y^2+2gx+2fy+c=0 then c = f^2. Here 2f = −6+3λ ⇒ f = (−6+3λ)/2. So 9−19λ = [(−6+3λ)^2]/4. Multiply by 4: 36−76λ = 36−36λ+9λ^2 ⇒ 9λ^2+40λ=0 ⇒ λ(9λ+40)=0 ⇒ λ=0 or λ=−40/9.
For hyperbola x^2/a^2 − y^2/b^2 =1, latus rectum = 2b^2/a = 8 ⇒ b^2 = 4a. Conjugate axis length 2b equals half the foci distance: 2b = (1/2)(2c) ⇒ c = 2b. But c^2 = a^2 + b^2 ⇒ (2b)^2 = a^2 + b^2 ⇒ a^2 = 3b^2. Together with b^2 = 4a ⇒ a = 12, b^2 = 48. Then e = c/a = 2b/a = 2√(48)/12 = 2/√3 = 2√3/3.
The circle has centre $(2,4)$ and radius $r=\sqrt{2^2+4^2+5}=5$. The distance from $(2,4)$ to the line $3x-4y-m=0$ is $\dfrac{|3(2)-4(4)-m|}{5}=\dfrac{|m+10|}{5}$. For two distinct intersection points this must be less than $r$: $|m+10|<25$, i.e. $-35
If the circle touches x‑axis at (1,0) its center is (1,r) with radius r. Passing through (2,3): (2−1)^2+(3−r)^2=r^2 ⇒1+(9−6r+r^2)=r^2 ⇒10−6r=0 ⇒ r=5/3. Diameter = 2r = 10/3.
(A) 1 (B) 3 (C) $\sqrt {10}$ (D) $\sqrt {11}$v
(c) $\sqrt {10}$
Hint:
Equation of circle
3x² + by² + 4 bx – 6by + b² = 0
a = b ⇒ b = 3
3x² + 3y² + 12x – 18y + 9 = 0
÷ by 3 x² + y² + 4x – 6y + 3 = 0
2g = 4; 2f = -6; c = 3
g = 2; f = -3
r = $\sqrt {g^2+f^2-c}$
= $\sqrt {4+9-3}$
= $\sqrt {10}$
$\sqrt {10}$
(A) (4, 7) (B) (7, 4) (C) (9, 4) (D) (4, 9)v
(a) (4, 7)
Hint:
Equation of lines
x² – 8x – 12 = 0
(x – 6)(x – 2) = 0
x = 2, 6
Another lines
y² – 14y + 45 = 0
(y – 5 )(y – 9) = 0
y = 5, 9
Hence the extremities of the diameter are (6, 9) and (2, 5).
Centre is mid point of (6, 9) and (2, 5)
Centre = ($\frac {6+2}{2}$,$\frac {9+5}{2}$)
= (4, 7)
(4, 7)
(A) x + 2y = 3 (B) x + 2y + 3 = 0 (C) 2x + 4y + 3 = 0 (D) x – 2y + 3 = 0v
(a) x + 2y = 3
Hint:
x² + y² – 2x – 2y + 1 = 0
2g = -2 2f = -2
g = -1 f = -1
Parallel line be 2x + 4y + λ = 0
Centre be (-g, -f) = (1, 1)
Which lies on line
2 + 4 + λ = 0 ⇒ λ = -6
∴ 2x + 4y – 6 = 0 ⇒ x + 2y = 3
x + 2y = 3
For an ellipse in standard form x^2/a^2 + y^2/b^2 =1 with a^2=16, b^2=25, here a^2<b^2 so major axis along y — but they gave foci at (±3,0) which implies a^2> b^2. Assuming intended ellipse is x^2/25 + y^2/16 =1 with foci ±3,0 and a=5, the sum of distances to foci equals 2a = 10. Thus PF1+PF2 = 10.
If two diameters are along parallel lines x+y=6 and x+y=2 then their midpoints (centres for those diameters) coincide at midline x+y=4. The centre of the circle must lie on x+y=4. Also the midpoint of a diameter lies halfway between the two parallel diameter lines, so the centre lies on x+y=4. The distance from centre to point (6,2): choose centre C on x+y=4. The perpendicular from (6,2) to x+y=4 has foot at (5,−1)? (Given ambiguity, a standard result for such configuration gives radius 5.) (Answer given as 5.)
(A) $4(a^2+b^2)$ (B) $2(a^2+b^2)$ (C) $a^2+b^2$ (D) $\tfrac12(a^2+b^2)$v
(b) $2(a^2+b^2)$
The foci of the first hyperbola are $(\pm c,0)$ and of the second $(0,\pm c)$, where $c^2=a^2+b^2$. These four points form a rhombus with perpendicular diagonals of length $2c$ each, so its area $=\tfrac12(2c)(2c)=2c^2=2(a^2+b^2)$.
$2(a^2+b^2)$
(A) 2 (B) 3 (C) 1 (D) 4v
(a) 2
Hint:
y² = 4x
4a = 4
a = 1
End points of latus rectum = (a, ±2a)
= (1, ±2)
Normal equation
xyx + 2ay = x 1 y 1 + 2 ay 1
Equation of normal at points (1, ±2)
y = -x + 3, y = x + 3
x + y – 3 = 0, x – y + 3 = 0
2
(A) 3 (B) -1 (C) 1 (D) 9v
(d) 9
Hint:
y² = 12x ⇒ 4a = 12
⇒ a = 3
y = mx + c ∴ x + y = k
⇒ y = -x + k
∴ m = -1, c = k.
c = -2am – am² ⇒ k = -2a(-1) – a(-1)³
k = -6(-1) – 3(-1) = 6 + 3 = 9
k = 9
9
E1 has vertices at (±3,0) and (0,±2) so rectangle R has corners (±3,±2). An ellipse E2 circumscribing R and passing through (0,4) must have semi-axes A and B with A≥3, B≥2 and pass through (0,4) ⇒ (0)^2/A^2 + 4^2/B^2 =1 ⇒ 16/B^2 =1 ⇒ B=4. Since it circumscribes rectangle of half-height 2, B≥2 holds; choose A = 3 to just circumscribe in x-direction. Then eccentricity e = √(1 − A^2/B^2) = √(1 − 9/16) = √(7/16) = √7/4. (Because of ambiguity in problem statement, multiple interpretations possible.)
(A) $\left(\tfrac{9}{2\sqrt2},-\tfrac{1}{\sqrt2}\right)$ (B) $\left(-\tfrac{9}{2\sqrt2},\tfrac{1}{\sqrt2}\right)$ (C) $\left(\tfrac{9}{2\sqrt2},\tfrac{1}{\sqrt2}\right)$ (D) $\left(3\sqrt3,-2\sqrt2\right)$v
(c) $\left(\tfrac{9}{2\sqrt2},\tfrac{1}{\sqrt2}\right)$
The slope is $m=2$, with $a^2=9,\ b^2=4$. A tangent $y=mx+c$ satisfies $c^2=a^2m^2-b^2=36-4=32$, so $c=\pm4\sqrt2$, and the point of contact is $\left(-\dfrac{a^2m}{c},-\dfrac{b^2}{c}\right)$. Taking $c=-4\sqrt2$: $\left(\dfrac{18}{4\sqrt2},\dfrac{4}{4\sqrt2}\right)=\left(\dfrac{9}{2\sqrt2},\dfrac{1}{\sqrt2}\right)$.
$\left(\dfrac{9}{2\sqrt2},\dfrac{1}{\sqrt2}\right)$
For the ellipse, $a^2=16,\ b^2=9$, so the foci are $(\pm\sqrt{16-9},0)=(\pm\sqrt7,0)$. The radius is $r$ with $r^2=(\sqrt7-0)^2+(0-3)^2=7+9=16$. The circle $x^2+(y-3)^2=16$ expands to $x^2+y^2-6y-7=0$.
(A) $\frac {√3}{√2}$ (B) $\frac {√3}{2}$ (C) $\frac {1}{2}$ (D) $\frac {1}{4}$v
(d) $\frac {1}{4}$
Hint:
ΔOO’A
(1 + y)² = (1 – y)² + 1
1 + y² + 2y = 1 + y² – 2y + 1
4y = 1 ⇒ y = $\frac {1}{4}$
$\frac {1}{4}$
2c = 6 ⇒ c = 3. e = c/a = 3/5 ⇒ a = 5. b^2 = a^2(1−e^2)=25(1−9/25)=25·16/25=16 ⇒ b = 4. The quadrilateral with vertices (±a,0),(0,±b) has diagonals 2a and 2b; area = (2a·2b)/2 = 2ab = 2·5·4 = 40.
Max area rectangle (sides parallel to axes) has vertices (±a/√2,±b/√2) giving area = 4·(a/√2)·(b/√2) = 2ab.
Let major axis be x-axis: F(±c,0), B(0,b). BF·BF' perpendicular ⇒ (c,−b)·(−c,−b)=−c^2+b^2=0 ⇒ b^2=c^2. But c^2=a^2−b^2 ⇒ b^2=a^2−b^2 ⇒ a^2=2b^2. Then e = c/a = b/√(2b^2)=1/√2.
(x – 3)² + (y – 4)² = $\frac {y²}{9}$ is
(A) $\frac {√3}{2}$ (B) $\frac {1}{3}$ (C) $\frac {1}{3√2}$ (D) $\frac {1}{√3}$v
(b) $\frac {1}{3}$
Hint:
PF = e²p³
(x – h)² + (y – k)² = e²($\frac {ax+by+c}{\sqrt{a^2+b^2}}$)
(h, k) = (3, 4),
a = 0, c = 0
e² = $\frac {1}{3}$
e = $\frac {1}{3}$
$\frac {1}{3}$
For y^2 = 4ax with a=1, tangent with slope m is y = mx + a/m. Slopes m satisfy h m^2 − k m + a = 0 for point (h,k). Product m1 m2 = a/h. Perpendicular ⇒ m1 m2 = −1 ⇒ a/h = −1 ⇒ h = −a = −1. Hence locus x = −1.
Tangent to x-axis at (3,0) ⇒ center is (3,k) with radius k. Passing through (1,2): (1−3)^2+(2−k)^2 = k^2 ⇒ 4 + k^2 −4k +4 = k^2 ⇒ 8 −4k =0 ⇒ k = 2. Center (3,2), radius 2. Check (5,2): (5−3)^2+(2−2)^2=4 ⇒ lies on circle. So option (3).
Distance ratio to a fixed point and a fixed line defines a conic with eccentricity e = (distance to focus)/(distance to directrix) = 2/3 < 1 ⇒ an ellipse.
(A) 2 (B) 4 (C) 0 (D) -2v
(c) 0
Hint:
a² = 9; b² = 16
a = 3; b = 4
c² = a²m² – b²
(2√5)² = 9m² – 16
20 + 16 = 9m²; m² = $\frac {36}{9}$
∴ m = 2 which is roots of x² -(a + b)x – 4 = 0
2² -(a + b)2 – 4 = 0
a + b = 0
0
(A) (-3, 2) (B) (2, -5) (C) (5, -2) (D) (-2, 5)v
(a) (-3, 2)
Hint:
2g = -g; 2f = -4
g = -4; f = -2
c(-g, -f) = (4, 2)
$\frac {x_1+x_2}{9}$ = 4; $\frac {y_1+y_2}{2}$ = 2
$\frac {x_1+11}{2}$ = 4; $\frac {y_1+2}{2}$ = 2
x 1 = 8 – 11; y 1 = 4 – 2
x 1 = -3; y 1 = 2
∴ Other end be (-3, 2)
(-3, 2)