Conic Sections - Class 11 Engineering Maths - Extra Questions

We have,

$$ \left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,0 \right) $$

$$ \left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,1 \right) $$

Then,

Equation of circle is

$$ \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0 $$

$$ \Rightarrow \left( x-1 \right)\left( x-0 \right)+\left( y-0 \right)\left( y-1 \right)=0 $$

$$ \Rightarrow x\left( x-1 \right)+y\left( y-1 \right)=0 $$

$$ \Rightarrow {{x}^{2}}-x+{{y}^{2}}-y=0 $$

$$ \Rightarrow {{x}^{2}}+{{y}^{2}}-x-y=0 $$

Equation of a conic in polar form is 

$$\dfrac { l }{ r } =1-e\cos { \theta  }$$

Let the eccentricities of the conies be $${ e }_{ 1 },{ e }_{ 2 },{ e }_{ 3 }....$$ and latustrecum $${ l }_{ 1 },l_{ 2 },{ l }_{ 3 }....$$ . Let the axes be inclined at angle $${ \alpha  }_{ 1 },\alpha _{ 2 },{ \alpha  }_{ 3 }....$$ then

$$\dfrac { l }{ { SP }_{ 1 } } =1-{ e }_{ 1 }\cos { (\theta -{ \alpha  }_{ 1 }) } \\ \Rightarrow \dfrac { 1 }{ { SP }_{ 1 } } =\dfrac { 1-{ e }_{ 1 }\cos { (\theta -{ \alpha  }_{ 1 }) }  }{ l } $$

Similarly $$\dfrac { l }{ { SP }_{ 2 } } =1-{ e }_{ 2 }\cos { (\theta -{ \alpha  }_{ 2 }) } $$

$$\Rightarrow \dfrac { 1 }{ { SP }_{ 2 } } =\dfrac { 1-{ e }_{ 2 }\cos { (\theta -{ \alpha  }_{ 2 }) }  }{ l } $$

Given $$\dfrac { 1 }{ SQ } =\dfrac { 1 }{ { SP }_{ 1 } } +\dfrac { 1 }{ { SP }_{ 2 } } ......$$

$$\Rightarrow \dfrac { 1 }{ SQ } =\dfrac { 1-{ e }_{ 1 }\cos { (\theta -{ \alpha  }_{ 1 }) }  }{ { l }_{ 1 } } +\dfrac { 1-{ e }_{ 2 }\cos { (\theta -{ \alpha  }_{ 2 }) }  }{ { l }_{ 2 } } ......\\ \Rightarrow \dfrac { 1 }{ SQ } =\left( \dfrac { 1 }{ { l }_{ 1 } } +\dfrac { 1 }{ { l }_{ 2 } } ..... \right) -\left( \dfrac { { e }_{ 1 } }{ { l }_{ 1 } } \cos { (\theta -{ \alpha  }_{ 1 }) } +\dfrac { { e }_{ 2 } }{ { l }_{ 2 } } \cos { (\theta -{ \alpha  }_{ 2 }) } ...... \right) $$

This is equation of form $$\dfrac { 1 }{ r } =\dfrac { 1 }{ l } -\dfrac { e }{ l } \cos { (\theta -\alpha ) } $$

Hence it represents a conic with focus $$S'$$ and latusrectum $$L$$

where $$\dfrac { 1 }{ L } =\dfrac { 1 }{ { l }_{ 1 } } +\dfrac { 1 }{ { l }_{ 2 } } +\dfrac { 1 }{ { l }_{ 3 } } .....$$

Hence proved

                $$\Rightarrow \text{L.R} = \dfrac{{2{b^2}}}{a} = {b}$$

                $$ \Rightarrow \dfrac{b}{a} = \dfrac{1}{2}$$              $$\dots(1)$$

$$\textbf{Step-2: Eccentricity of ellipse is given as-}$$

                 $$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}} $$ $$\dots(2)$$

$$\textbf{Step-3: Putting values (1) in (2)}$$

                  $$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}}$$

                     $$ = \sqrt {1 - {{\left( {\dfrac{1}{2}} \right)}^2}}  $$

                     $$= \sqrt {1 - \dfrac{1}{{4}}} $$

                     $$ = \sqrt {\dfrac{{3}}{{4}}}$$

$$\textbf{ So, the eccentricity of ellipse is given by}$$ $$ \mathbf{{\dfrac{{\sqrt3}}{{2}}}}. $$

$${\mathbf{Step}}\;1:\;{\mathbf{Finding}}\;{\mathbf{the}}\;{\mathbf{radius}}\;{\mathbf{of}}\;{\mathbf{the}}\;{\mathbf{circle}}$$

                  $${\text{Centre}}: \left( {2,2} \right)$$

                  $${\text{Circle passes through }}\left( {4,5} \right).$$

                  $$\Rightarrow {\text{Radius of the circle}} = {\text{ Distance between }}\left( {2,2} \right){\text{ and }}\left( {4,5} \right).$$

                  $${\text{Therefore}},$$

                 $$ =\sqrt{ (4-2)^2 + (5-2)^2 }$$

                 $$ = \sqrt{ (4 + 9)} $$

                 $$ = \sqrt {13}$$

$${\mathbf{Step}}\;2 :{\mathbf{Finding}}\;{\mathbf{the}}\;{\mathbf{equation}}\;{\mathbf{of}}\;{\mathbf{the}}\;{\mathbf{circle}}$$

                $${\text{We know}},$$

                $${\text{ The equation of the circle is}}:$$

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

                $${\left( {x - 2} \right)^2} + {\left( {{\text{y}} - 2} \right)^2}= {\left( {\sqrt {13} } \right)^2}$$

                $${{\text{x}}^2} - \text{4x +4} + {{\text{y}}^2} - 4{\text{y}} + 4{\text{ }} =   13$$

                $${{\text{x}}^2} + {{\text{y}}^2} -\text{4x} - 4{\text{ y }} - 5 = 0$$

$${\mathbf{Hence}},{\mathbf{the}}\;{\mathbf{equation}}\;{\mathbf{of}}\;{\mathbf{the}}\;{\mathbf{circle}}\;{\mathbf{is}}\;{{\boldsymbol{ x^2 + y^2 - 4x - 4y -5 = 0}}}$$

We know that the equation of a circle with center at $$\left( h,k \right)$$ is,

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

Now, this circle passes through $$\left( 2,-2 \right)$$. Then

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

$$ {{h}^{2}}+{{k}^{2}}-4h+4k+8={{r}^{2}} $$

Again, this circle passes through $$\left( 3,4 \right)$$. Then

$$ {{\left( 3-h \right)}^{2}}+{{\left( 4-k \right)}^{2}}={{r}^{2}} $$

$$ {{h}^{2}}+{{k}^{2}}-6h-8k+25={{r}^{2}} $$

Subtracting the above results, we have

$$2h+12k=17$$     …… (1)

Now, the center lies on $$x+y=2$$. So,

$$h+k=2$$     ….. (2)

Solving equations (1) and (2), we get

$$h=\dfrac{7}{10}$$ and $$k=\dfrac{13}{10}$$

Put values of $$\left( h,k \right)$$ in the first expression.

$$ {{\left( 2-\dfrac{7}{10} \right)}^{2}}+{{\left( -2-\dfrac{13}{10} \right)}^{2}}={{r}^{2}} $$

$$ {{r}^{2}}=\dfrac{169+1089}{100} $$

$$ r=\dfrac{\sqrt{1258}}{10} $$

Therefore, equation of the circle is,

$${{\left( x-\dfrac{7}{10} \right)}^{2}}+{{\left( y-\dfrac{13}{10} \right)}^{2}}=\dfrac{1258}{100}$$

Hence, this is the required equation.

We have,

A variable circle passes through the point $$A\left( 2,1 \right)$$and touches the X-axis.

So,

In a circle AB is a diameter where the coordinate of point A is (2,1) and let the coordinate of B is $$\left( {{x}_{1}},{{y}_{1}} \right)$$.

Now, we know that,

Equation of circle in diameter from is

$$\left( x-{{x}_{1}} \right)\left( x-2 \right)+\left( y-{{y}_{1}} \right)\left( y-1 \right)=0$$

And it touches X-axis so $$y=0$$

$$ \left( x-{{x}_{1}} \right)\left( x-2 \right)+\left( 0-{{y}_{1}} \right)\left( 0-1 \right)=0 $$

$$ \Rightarrow {{x}^{2}}-x{{x}_{1}}-2x+2{{x}_{1}}+{{y}_{1}}=0 $$

$$ \Rightarrow {{x}^{2}}+\left( -2-{{x}_{1}} \right)x+2{{x}_{1}}+{{y}_{1}}=0 $$

Also, the discriminate of above equation will be equal to zero,

Because circle touches X-axis.

$$ {{\left( -2-{{x}_{1}} \right)}^{2}}=4\left( 2{{x}_{1}}+{{y}_{1}} \right) $$

$$ \Rightarrow 4+4{{x}_{1}}+{{x}_{1}}^{2}=8{{x}_{1}}+4{{y}_{1}} $$

$$ \Rightarrow 4-4{{x}_{1}}+{{x}_{1}}^{2}=4{{y}_{1}}^{2} $$

$$ \Rightarrow {{\left( {{x}_{1}}-2 \right)}^{2}}=4{{y}_{1}}^{2} $$

Hence, the locus of end pint of diameter.

We have,

Equation of line is $$4x-3y+2=0\,\,\,.......\,\,\left( 1 \right)$$

Now,

The length of perpendicular from $$\left( 3,-2 \right)$$ to given line $$4x-3y+2=0$$

So,

$$\text{Length}\,\,=\,\,P\,\,=\,\left| \dfrac{4x-3y+2}{\sqrt{{{4}^{2}}+{{\left( -3 \right)}^{2}}}} \right|$$

At point$$\left( 3,-2 \right)$$

So,

$$ P=\left| \dfrac{4\times 3-3\times \left( -2 \right)+2}{5} \right| $$

$$ P=\dfrac{20}{5}=4\,\,unit\, $$

Now, As circle  cuts off an intercepts of length 6,

So, it intercepts at a distance of 3 on each so, from foot of perpendicular.

Hence, the radius of circle is

$$ =\sqrt{{{4}^{2}}+{{3}^{2}}} $$

$$ \sqrt{16+9}=\sqrt{25}=5 $$

Hence, equation of circle is

$$ {{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=5 $$

$$ or\,\,{{x}^{2}}+9-6x+{{y}^{2}}+4+4y=5 $$

$$ \Rightarrow {{x}^{2}}+{{y}^{2}}-6x+4y+13-5=0 $$

$$ \Rightarrow {{x}^{2}}+{{y}^{2}}-6x+4y+8=0 $$

Hence, this is the answer.

We have,

Line $$4x-3y+4=0$$….. (1)

Line $$x-y-1=0$$  ……  (2)

From equation (1) and (2) to and we get,

$$x=-7\,and\,y=-8$$

Then,

Centre of circle $$\left( h,k \right)=\left( -7,-8 \right)$$

But centre lies third quadrant.

So,

centre$$\left( h,k \right)=\left( -7,8 \right)$$

so,

Equation of circle

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

$$ \Rightarrow {{\left( x+7 \right)}^{2}}+{{\left( y-8 \right)}^{2}}={{\left( -7 \right)}^{2}}\,\,\,\left( \therefore \,x-axis\,so,\,radius\,of\,circle\,=Abssicissa \right) $$

$$ \Rightarrow {{x}^{2}}+49+14x+{{y}^{2}}+64-16y=49 $$

$$ \Rightarrow {{x}^{2}}+{{y}^{2}}+14x-16y+64=0 $$

Hence, the required equation of circle.

take the parabola as $$y^2=4ax$$

 now coordinates of its end of latus rectum are $$(a,2a) ; (a,-2a)$$.

Now tangent at any point $$(x`,y`)$$ on parabola is $$yy`=2a(x+x`)$$.

Tangent at point $$(a,2a)$$ is $$2ay=2ax+2a^2$$ and at $$(a,-2a)$$ is $$-2ay=2ax+2a^2 $$.

By solving this linear equation in two unknown, we get point of intersection of tangent $$y=0$$ and $$x=-a$$;

so the point of intersection is $$(-a,0)$$ and it clearly lie on directrix. 

We have,

Equation of parabola

$${{y}^{2}}=4ax$$

Focus has positive x co ordinate.

So,

Co ordinate of focus $$=\left( a,0 \right)$$

$$\left( a,0 \right)=\left( 6,0 \right)$$

$$a=6$$

Hence, equation of parabola is

$$ {{y}^{2}}=4ax $$

$$ {{y}^{2}}=4\times 6x $$

$$ {{y}^{2}}=24x $$

Then,

Latus rectum $$=4a=4\times 6=24$$

Given that we need to find the equation of the circle which circumscribes the square ABCD of side a.

It is also told that AB and AD are assumed as x and y - axes.

Assuming A as origin, We get the points $$B(a,0)\ and\ D(0,a)$$.

So, we need to find the circle which is passing through the points $$A(0,0), B(a,0)\ and\ D(0,a)$$.

We know that the standard form of the equation of the circle is given by:

$$\Rightarrow x^2+y^2+2fx+2gy+c=0$$   $$........(1)$$

$$Substituting\ A(0,0)\ in\ (1),\ we\ get,$$

$$0^2+0^2+2f(0)+2g(0)+c=0$$

$$\Rightarrow c=0$$  $$ ........(2)$$

$$Substituting\ B(a,0)\ in\ (1),\ we\ get,$$

$$a^2+0^2+2f(a)+2g(0)+c=0$$

$$\Rightarrow a^2+2fa=0$$  $$........(3)$$

$$Substituting\ D(0,a)\ in\ (1),\ we\ get,$$

$$0^2+a^2+2f(0)+2g(a)+c=0$$

$$\Rightarrow a^2+2ga=0$$  $$.......(4)$$

$$On\ solving\ (2),\ (3)\ and\ (4)\ we\ get,$$

⇒ 

$$Substituting\ these\ values\ in\ (1),\ we\ get$$

⇒ 

$$\Rightarrow x^2+y^2-ax-ay=0$$

$$\Rightarrow x^2+y^2-a(x+y)=0$$

∴Thus proved.