Maths M.C.Q (1 Marks)

$ = \frac{{( - 36)( - 14) - (23)( - 2)}}{{(2)(23) - {{(36)}^2}}},\;\frac{{( - 2)( - 36) - (2)( - 14)}}{{(2)(23) - {{( - 36)}^2}}}$

$ = \left( { - \frac{{11}}{{25}},\, - \frac{2}{{25}}} \right)$.

$i.e.$, $\left( {\frac{{hf - bg}}{{ab - {h^2}}},\,\frac{{gh - af}}{{ab - {h^2}}}} \right)$.

Then $\sqrt {{{(x - 1)}^2} + {{(y + 1)}^2}} = \sqrt 2 \left( {\frac{{x - y + 1}}{{\sqrt 2 }}} \right)$,

[Using $SP = e.PM]$

$ \Rightarrow 2xy - 4x + 4y + 1 = 0$.

Squaring both sides, we get $\sqrt {{{(x + 2)}^2} + {y^2}} = x + 2$

Again squaring both sides, we get ${y^2} = 0$, which is the equation of pair of straight lines.

Comparing the given equation with

$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$

We get, $a = 1$, $h = 0$, $b = 1$, $g = 1$, $f = 0$, $c = - 1$

$\therefore $ $\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$

$\Delta = 1 + 0 + 0 - 1 = 0$

Hence, the given equation represents two straight lines.

$ = 2 - \frac{9}{4} - 2 < 0$ and ${h^2} - ab = 1 - 1 = 0$.

i.e., ${h^2} = ab$ ==> a parabola.

Similarly eliminating s from $x = 2s,\,y = \frac{2}{s},$

we get $xy = 4.$

Hence point of intersection is $(2, 2).$

$\frac{{{{(x - 1)}^2}}}{2} + {\left( {y + \frac{3}{4}} \right)^2} = \frac{1}{{16}}$

==> $\frac{{{{(x - 1)}^2}}}{{(1/8)}} + \frac{{{{\left( {y + \frac{3}{4}} \right)}^2}}}{{(1/16)}} = 1$,

which is an ellipse with ${a^2} = \frac{1}{8}$ and ${b^2} = \frac{1}{{16}}$

$\therefore \frac{1}{{16}} = \frac{1}{8}(1 - {e^2})$

==>${e^2} = 1 - \frac{1}{2}$

==> $e = \frac{1}{{\sqrt 2 }}$.

$ \Rightarrow $ ${(2x - 6)^2} + {(4y - 4)^2} = 53$

$ \Rightarrow $ $4\,{(x - 3)^2} + 16\,{(y - 1)^2} = 53$

$ \Rightarrow $ $\frac{{{{(x - 3)}^2}}}{{53/4}} + \frac{{{{(y - 1)}^2}}}{{53/16}} = 1$

$\therefore $ $e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \frac{{53/16}}{{53/4}}}$

$= \sqrt {1 - \frac{1}{4}} = \frac{{\sqrt 3 }}{2}$.

Focus $S( - 2,\,0)$,

Equation of directrix $2x - 9 = 0$

${(PS)^2} = \frac{4}{9}{(PM)^2}$

==> ${(h + 2)^2} + {(k)^2} = \frac{4}{9}{\left( {\frac{{2h - 9}}{2}} \right)^2}$

==> $9[{(h + 2)^2} + {(k)^2}] = \frac{{4{{(2h - 9)}^2}}}{4}$

==> $9{h^2} + 9{k^2} + 36h + 36 = 4{h^2} + 81 + 36h$

==> $\frac{{5{h^2}}}{{45}} + \frac{{9{k^2}}}{{45}} = 1$

==> $\frac{{{h^2}}}{9} + \frac{{{k^2}}}{5} = 1$ 

Locus of point $P(h, k)$ is $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$, which is an ellipse

$i.e.$, a hyperbola.

$\Delta = (1)\,(1)\,(2) + 2(2)\,(1)\,(2) - (1)\,{(2)^2} - (1)\,{(1)^2} - 2{(2)^2} < 0$

Hence it is a hyperbola.

$⇒$  $ - \frac{{{{(x + 1)}^2}}}{4} + \frac{{{y^2}}}{4} = 1$

Equation of directrices of $\frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1$ are $y = \pm \frac{b}{e}$

Here $b = 2,\;e = \sqrt {1 + 1} = \sqrt 2 $

Hence $y = \pm \frac{2}{{\sqrt 2 }}$

$⇒$ $y = \pm \sqrt 2 $.

Now let the various point be $(h,k)$,

then accordingly $\frac{{\sqrt {{{(h - 1)}^2} + {{(k - 1)}^2}} }}{{\frac{{2h + k - 1}}{{\sqrt 5 }}}} = \sqrt 3 $

Squaring both the sides, we get

$5[{(h - 1)^2} + {(k - 1)^2}] = 3{(2h + k - 1)^2}$

On simplification, the required locus is

$7{x^2} + 12xy - 2{y^2} - 2x + 4y - 7 = 0$.

==> $({x^2} - 2x) - 4({y^2} - 4y) - 40 = 0$

==> ${(x - 1)^2} - 1 - 4[{(y - 2)^2} - 4] - 40 = 0$

==> ${(x - 1)^2} - 4{(y - 2)^2} = 25$

==> $\frac{{{{(x - 1)}^2}}}{{25}} - \frac{{{{(y - 2)}^2}}}{{25/4}} = 1$, which is a hyperbola.

$h = \frac{{a\,\cos \alpha - b\,\sin \alpha }}{{\cos 2\alpha }}$, $k = \,\frac{{a\,\sin \alpha - \,b\cos \alpha }}{{\cos 2\alpha }}$

Then the locus of point $(h,\,k)$ is ${x^4} + {y^4} - 2{x^2}{y^2}$ $ = ({a^2} + {b^2})$ $({x^2} + {y^2}) + 4abxy$, which is not a locus of any given curves.

Obviously, it is a parabola whose focus is $\left( {\frac{5}{2},1} \right)$ and directrix

is $x = \frac{3}{2}$.

==> $9{y^2} = 4ax$.

Hence vertex is $(1, -1)$, which lies in $IV$ quadrant.

Hence the vertex is $\left( {\frac{1}{3},\, - \frac{2}{9}} \right)$.

Passes through $(3, 6)$ 

==> $16 = 4a.8$

==> $a = \frac{1}{2}$

==> ${(x + 1)^2} = 2(y + 2)$

==> ${x^2} + 2x - 2y - 3 = 0$.

$\frac{{\sqrt {{{(x + 8)}^2} + {{(y + 2)}^2}} }}{{\left| {\frac{{2x - y - 9}}{{\sqrt 5 }}} \right|}} = 1$

Squaring and after simplification, we get

${x^2} + 4{y^2} + 4xy + 116x + 2y + 259 = 0$.

Focus is $(-3, 0)$ 

$⇒$  $2a = ( - 5 + 3) = 2$ $⇒$  $a = 1$

Vertex is $\left( {\frac{{ - 3 + ( - 5)}}{2},\,0} \right) = ( - 4,\,0)$

Therefore, equation is ${(y - 0)^2} = 4(x + 4)$.

where $A = (a' - a)$ or ${y^2} = 4(a' - a)(x - a)$.

==> ${(y - 2)^2} = - 4(x - 1)$

Vertex is $(1,2)$ and focus = $(0,2)$.

==> ${(x + 2)^2} = - 2\left( {y - \frac{{11}}{2}} \right)$

Hence vertex is $\left( { - 2,\frac{{11}}{2}} \right)$.

So none is correct.

Equation of latus rectum $x + \lambda = 0$

Vertex $ = \left( {\frac{{ - C + {B^2}}}{{2A}},B} \right)$, focus $ \equiv \left( {\frac{{ - C + {B^2}}}{{2A}} - \frac{A}{2},B} \right)$

Equation of $L.R.$ is $x = \frac{{ - C + {B^2}}}{{2A}} - \frac{A}{2} = \frac{{{B^2} - {A^2} - C}}{{2A}}$.

$\therefore $ $a = 2$

Hence parabola is ${(x - 0)^2} = - 4.2(y - 4)$

$i.e.$,${x^2} + 8y = 32$.

==> ${(3x - 1)^2} = - 36y - 18 = - 36\left( {y + \frac{1}{2}} \right)$

==> $9{\left( {x - \frac{1}{3}} \right)^2} = - 36\left( {y + \frac{1}{2}} \right)$

Hence length of latus rectum is $4$.

==> ${\left( {y - \frac{2}{3}} \right)^2} = \frac{{16}}{9}\left( {x + \frac{{61}}{{16}}} \right)$

Put $y - \frac{2}{3} = Y$and $x + \frac{{61}}{{16}} = X$

==> ${Y^2} = 4\left( {\frac{4}{9}} \right)\,X$

Axis of this parabola is $Y = 0$

==> $y - \frac{2}{3} = 0$

==> $3y = 2$.

where $x = X + a,\,\,y = Y + b$.

Therefore the equation of the parabola referred to the point $(a,b)$ as the vertex is

${(x - a)^2} = l(y - b)$ or ${(x - a)^2} = \frac{l}{2}(2y - 2b)$.

${(y - 2)^2} = 5(x + 1)$

Obviously, latus rectum is $5$.

or ${x^2} - 2x + 1 = 2y + 1$

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

Here $4a = 2 $

$\Rightarrow a = \frac{1}{2}$

Now focus is $\left( {x - 1 = 0,\,y + \frac{1}{2} = \frac{1}{2}} \right)\,\,i.e.,\,(1,\,0)$.

${y^2} - 4y - 2x - 8 = 0$

==> ${(y - 2)^2} = 2(x + 6)$

$\therefore $Length of latus rectum = $2$.

On solving we get

${(ax - by)^2} - 2{a^3}x - 2{b^3}y + {a^4} + {a^2}{b^2} + {b^4} = 0$.

$2{\left( {y - \frac{5}{2}} \right)^2} = - (x - 4)$

==> $4a = \frac{1}{2}$.

${(y + 1)^2} = \frac{3}{2}(x + 3).$

So, vertex is $( - 3, - 1)$.

$4{\left( {y - \frac{1}{2}} \right)^2} = 6(x + 1)$

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

$\Rightarrow {Y^2} = \frac{3}{2}X$

where, $Y = y - \frac{1}{2},\,\,X = x + 1$

$\therefore y = Y + \frac{1}{2},\,\,\,x = X - 1$…..$(i)$

For focus $X = a,\,\,Y = 0$

$\because 4a = \frac{3}{2} \Rightarrow a = \frac{3}{8} \Rightarrow x = \frac{3}{8} - 1 =  - \frac{5}{8}$

$y = 0 + \frac{1}{2} = \frac{1}{2}$, Focus$ = \left( { - \frac{5}{8},\frac{1}{2}} \right)$.

It can be written as ${(x + 4)^2} = - 12y + 12$

$ \Rightarrow \,{(x + 4)^2} = - 12(y - 1)$,  vertex is $( - \,4,\,1).$

Comparing the given equation with the standard equation, we get vertex $(\alpha ,\,\beta ) = ( - 3,\,2)$ and $a = 5.$

Therefore focus of the parabola $(\alpha + a,\,\beta ) = (2,\,2)$.

$ \Rightarrow $${x^2} - 4x = 8y - 12$

$ \Rightarrow $${(x - 2)^2} = 8(y - 1)$

Hence the length of latus rectum =$4a = 8$.

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

$ \Rightarrow \,{x^2} + \frac{x}{2} = \frac{y}{2}$

$ \Rightarrow \,{\left( {x + \frac{1}{4}} \right)^2} = \frac{y}{2} + \frac{1}{{16}}$

$ \Rightarrow \,{\left( {x + \frac{1}{4}} \right)^2} = \frac{1}{2}\left( {y + \frac{1}{8}} \right)$

It can be written as, ${X^2} = \frac{1}{2}Y$.....$(i)$

Here $A = \frac{1}{8}$, focus of $(i)$ is $\left( {0,\frac{1}{8}} \right)$

$i.e.$ $X = 0$, $Y = \frac{1}{8}$

==> $x + \frac{1}{4} = 0$, $y + \frac{1}{8} = \frac{1}{8}$

$ \Rightarrow \,x = - \frac{1}{4},$ $y = 0$

$i.e.$ focus of given parabola is $\left( { - \frac{1}{4},\,0} \right)$.