Conic Sections Test - 68

As we know,

Standard Equation of ellipse \(=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Length of latus rectum \(=\frac{2 b ^{2} }{a}\), when \(a > b\) and \(\frac{2 a ^{2} }{b}\), when \(a < b\).

Given,

\(3 x^{2}+y^{2}-12 x+2 y+1=0 \)

\(\Rightarrow 3\left(x^{2}-4 x+4\right)-12+\left(y^{2}+2 y+1\right)=0 \)

\(\Rightarrow 3(x-2)^{2}-12+(y+1)^{2}=0 \)

\(\Rightarrow 3(x-2)^{2}+(y+1)^{2}=12 \)

\(\Rightarrow \frac{3(x-2)^{2}}{12}+\frac{(y+1)^{2}}{12}=1 \) (Divide by 12)

\(\Rightarrow \frac{(x-2)^{2}}{4}+\frac{(y+1)^{2}}{12}=1 \)

\(\Rightarrow \frac{(x-2)^{2}}{2^{2}}+\frac{(y+1)^{2}}{(2 \sqrt{3})^{2}}=1 \)

\(\therefore a^{2}=2^{2}\) and \(b^{2}=(2 \sqrt{3})^{2}\)

Here \(a < b\)

So, length of latus rectum \(=\frac{2 a ^{2} }{ b}\)

\(=\frac{2(4)}{2 \sqrt{3}}\)

\(=\frac{4}{\sqrt{3}}\) units

As we know that,

Length of latus rectum of a hyperbola is given by,

\(\frac{2 b^{2}}{a}\)

Given,

Eccentricity \(=3\)

Length of latus rectum \(=4\)

\(\Rightarrow \frac{2 b^{2}}{a}=4 \)

\(\Rightarrow b^{2}=2 a\)

As we know that,

The eccentricity of a hyperbola is given by,

\(e=\frac{\sqrt{a^{2}+b^{2}}}{a}\)

\(\Rightarrow a^{2} e^{2}=a^{2}+b^{2} \)

\(\Rightarrow (3)^2 a^{2}=a^{2}+2 a \)

\(\Rightarrow 9 a^{2}=a^{2}+2 a \)

\(\Rightarrow 9 a^{2}-a^{2}=2 a \)

\(\Rightarrow 8a^{2}=2a\)

\(\Rightarrow a=\frac{1 }{ 4} \)

\(\Rightarrow a^2=\frac{1}{16} \)

\(\because b^{2}=2 a \)

\(\Rightarrow b^{2}=\frac{1 }{ 2}\)

As we know,

The equation of the required hyperbola is given by,

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

\(\Rightarrow \frac{x^{2}}{\frac{1}{16}}-\frac{y^{2}}{\frac{1}{2}}=1\)

\(\Rightarrow 16 x^{2}-2 y^{2}=1\)

So, the equation of the required hyperbola is \(16 x^{2}-2 y^{2}=1\).

As we know,

The general equation of a non-degenerate conic section is \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) where \(A, B\) and \(C\) are all not zero.

The above-given equation represents a non-degenerate conics whose nature is given below in the table:

As we know,

Equation of ellipse \(=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Foci \(=(\pm a e, 0)\)

Eccentricity \((e)=\sqrt{1-\frac{ b ^{2}}{ a ^{2}}}\)

Here, foci \(=(\pm 2,0)=(\pm a e, 0)\) and the eccentricity, \(e=\frac{1 }{ 2}\)

\(ae =2\)

\(\Rightarrow a \times \frac{1 }{ 2}=2 \)

\(\Rightarrow a =4 \)

\(\Rightarrow a ^{2}=16\)

Now, \(e=\sqrt{1-\frac{b^{2}}{a^{2}}}\)

\(\Rightarrow b^{2}=a^{2}\left(1-e^{2}\right)\)

\(\Rightarrow b^{2}=16(1-\frac{1 }{ 4}) \)

\(\Rightarrow b^{2}=16 \times(\frac{3 }{ 4}) \)

\(\Rightarrow b^{2}=12\)

\(\therefore\) Equation of ellipse is \(\frac{ x ^{2}}{16}+\frac{ y ^{2}}{12}=1\)

As we know,

The equation of a circle is given by:

\(\left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2}\)

Where \(\left( h , k \right)\) is the center of the circle and \(r\) is the radius.

Given,

\(x=2+3 \cos \theta \)

\(\Rightarrow 3 \cos \theta=x-2 \)..(i)

Also \( y=1-3 \sin \theta \)

\(\Rightarrow 3 \sin \theta=-y+1 \)...(ii)

Squaring adding (i) & (ii), we get

\(\left( x -2\right)^{2}+\left(-( y -1)\right)^{2}=3^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\)

\(\Rightarrow( x -2)^{2}+\left(-( y -1)\right)^{2}=3^{2}\)

Comparing it with standard equation of a circle, we get

Centre \(=\left(h, k\right)=\left(2,1\right)\) and radius \(=3\)

As we know,

If the point lies inside the circle no tangent can be drawn.

If the point lies on the circle then only one tangent can be drawn.

If the point lies outside the circle then a maximum of two tangent lines can be drawn on the circle.

Given,

Point \(=(2,6)\)

Equation of circle is \(x^{2}+y^{2}=40\)

\(x^{2}+y^{2}-40=0\)

Substituting \((2,6)\) in this equation, we get

\(2^{2}+6^{2}-40=0\)

\(\Rightarrow 4+36-40=0\)

\(\Rightarrow40-40=0\)

So, the point \((2,6)\) lies on the circle.

Therefore, only one tangent can be drawn.

As we know,

The standard equation of a hyperbola:

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)

where \(2 a\) and \(2 b\) are the length of the transverse axis and conjugate axis respectively and centre \((h, k)\).

The center is the midpoint of the 2 foci.

The eccentricity \(=\frac{\sqrt{a^{2}+b^{2}}}{a}\)

Length of latus recta \(=\frac{2 b ^{2}}{ a }\)

Distance from center to focus \(=\sqrt{a^{2}+b^{2}}\)

Given,

Foci are \((5,0)\) and \((-3,0)\).

Center \(=\left(\frac{5+(-3)}{2}, \frac{0+0}{2}\right)=(1,0)\)

Now distance of focus from the center \(=\sqrt{a^{2}+b^{2}}\)

\(\Rightarrow \sqrt{(5-1)^{2}+(0-0)^{2}}=\sqrt{ a ^{2}+ b ^{2}} \)

\(\Rightarrow 16=a ^{2}+ b ^{2}\)...(i)

Eccentricity \(=2\)

\(\therefore \frac{\sqrt{a ^{2}+ b^{2}}}{ a }=2\)

\(\Rightarrow \frac{\sqrt{16}}{ a }=2\)

\(\Rightarrow \frac{4}{ a }=2\)

\(\Rightarrow a =2\)

\(\Rightarrow a^{2}=4\)

Putting the value of \(a\) in (i), we get

\( 4+b^{2}=16 \)

\(\Rightarrow b^{2}=12\)

The equation of the hyperbola is given by:

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \)

\(\Rightarrow \frac{(x-1)^{2}}{4}-\frac{(y-0)^{2}}{12}=1 \)

\(\Rightarrow \frac{(x-1)^{2}}{4}-\frac{y^{2}}{12}=1\)

As we know,

The general equation of a non-degenerate conic section is \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) where \(a, h\) and \(b\) are all not zero

The above-given equation represents a non-degenerate conics whose nature is given below in the table: