Conic Sections Test - 67

As we know,

Identification of curves represented by the general equation of the second degree:

Let a general equation of second degree in \(x\) and \(y\) be \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\)...(1).

Then this equation represents,

1. A parabola if \(\Delta \neq 0\) and \(h^{2}=a b\)

2. An ellipse if \(\Delta \neq 0\) and \(h^{2}>a b\)

3. A hyperbola if \(Δ ≠ 0\) and \(h^2 > ab\)

4. A pair of straight line or empty set if \(\Delta=0\) and \(h^{2} \geq a b\)

5. A unique point if \(\Delta=0\) and \(h^2 < ab\)

6. A circle if \(a = b ≠ 0, h = 0\) and \(g^2 + f^2 -ac > 0\)

The given equation is \(2 x^{2}+2 y^{2}-5 x-7 y-3=0\).

Comparing it wth the equation (1) we get,

\(a=2, h=0, b=2, g=-(\frac{5 }{ 2}), f=-(\frac{7 }{ 2}), c=-3\)

As, \(a=b=2(>0), h=0\)

As we know,

\(g^2 + f^2 -ac > 0\)

\(\therefore (-\frac{5 }{ 2})^2+ (-\frac{7 }{ 2})^2-(2 \times -3)>0\)

\(\Rightarrow \frac{25 }{4}+ \frac{49}{4}+6>0\)

\(\Rightarrow \frac{25 +49+24}{4}>0\)

\(\Rightarrow \frac{98}{4}>0\)

\(\Rightarrow 24.5>0\)

So, the given equation represents a circle.

As we know,

Equation of vertical hyperbola \(\frac{y^{2}}{a^2}-\frac{x^{2}}{b^2}=1\).

Length of the conjugate axis of a hyperbola \(=2 b\)

Given,

Equation of hyperbola is \(\frac{y^{2}}{16}-\frac{x^{2}}{49}=1\).

So, by comparing the given equation of hyperbola with \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) we get,

\( a^{2}=16\) and \(b^{2}=49\)

So, the length of the conjugate axis for the given hyperbola \(=2b=2 \times 7=14\) units

As we know,

Standard equation of a hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\).

Coordinates of foci are \((\pm a e, 0)\).

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

Given,

Coordinates of foci \(=(\pm 3,0)\) and eccentricity, \(e=\frac{3}{2}\)

So, \(ae =3\)

As \(e=\frac{3}{2}\)

\(\therefore a\times \frac{3}{2} =3\)

\(\Rightarrow a=\frac{3\times2}{3}\)

\(\Rightarrow a=2\)

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

Squaring both sides, we get

\(e^2=1+\frac{b^{2}}{a^{2}}\)

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

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

Putting the value of \(a\) and \(e\), we get

\( b^{2}=2^{2}\left[\left(\frac{3}{2}\right)^{2}-1\right]\)

\(\Rightarrow b^{2}=4 \left(\frac{9}{4}-1\right)\)

\(\Rightarrow b^{2}=4\left(\frac{9-4}{4}\right)\)

\(\Rightarrow b^{2}=4 \times \frac{5}{4}\)

\(\Rightarrow b^{2}=5\)

So, the equation of the required hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{5}=1\).

As we know,

The standard form of the equation of a circle is \(\left( x - h \right)^{2}+\left( y - k \right)^{2}= R ^{2}\).

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

Area of the circle \(=\pi R ^{2}\)

Given,

Area of circle \(=154\) sq.units

\(\Rightarrow \pi R ^{2}=154 \)

\(\Rightarrow R ^{2}=154 \times \frac{7}{22} \)

\(\Rightarrow R =7\)

Equation of the diameters:

\(2 x-3 y+12=0 \)...(i)

\(x+4 y-5=0\)...(ii)

Intersection of the diameters (i) - \(2 \times\) (ii), we get

\(-11 y+22=0 \)

\(\Rightarrow y = 2\)

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

\( 2 x -3(2)+12=0 \)

\(\Rightarrow 2 x =6\)

\( \Rightarrow x =- 3\)

The center will be \(\left(-3,2\right)\).

By the standard equation of circle

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

\(\Rightarrow( x -(-3))^{2}+( y -2)^{2}=7^{2}\)

\(\Rightarrow( x +3)^{2}+( y -2)^{2}=49\)

\(\Rightarrow x ^{2}+9+6 x + y ^{2}+4-4 y -49=0\)

\(\Rightarrow x ^{2}+6 x + y ^{2}-4 y - 3 6 = 0\)

As we know, Standard equation of a circle,

\((x-h)^{2}+(y-k)^{2}=R^{2}\)

Where the center is \((h, k)\) and radius is \(R\).

The distance between a point on a circle and the center is the radius of the circle.

Distance between 2 points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is:

\(D=\sqrt{\left(y_{2}-y_{1}\right)^{2}+\left(x_{2}-x_{1}\right)^{2}}\)

Perpendicular distance of a point \(\left(x_{1}, y_{1}\right)\) from the line \(a x+b y+c=0\),

\(D=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)

Given, The circle touches the x and y-axis, i.e., the x and y-axis are tangent to the circle.

Equation of the \(x\)-axis is \(y=0\).

Radius is perpendicular distance of centre \(\left(-2,-2\right)\) from the tangent \(\left(y=0\right)\)

Radius \( r=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right| \)

\(\Rightarrow r=\left|\frac{0 \times(-2)+1 \times(-2)+0}{\sqrt{0^{2}+1^{2}}}\right| \)

\(\Rightarrow r=|-2|=2\)

Equation of circle having centre \(\left(-2,-2\right)\) and radius \(r=2\) is:

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

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

\(\Rightarrow x^{2}+y^{2}+4 x+4 y+8=4 \)

\(\Rightarrow x^{2}+y^{2}+4 x+4 y+4=0\)