Locus MCQ Questions & Answers in Geometry | Maths

Given that $$a>2b>0$$    and $$m>0$$
Also $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = mx - b\sqrt {1 + {m^2}} .....(1)$$
is tangent to $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + {y^2} = {b^2}.....(2)$$
as well as to $${\left( {x - a} \right)^2} + {y^2} = {b^2}.....(3)$$
$$\because $$ (1) is tangent to (3)
$$\therefore \left| {\frac{{am - b\sqrt {1 + {m^2}} }}{{\sqrt {{m^2} + 1} }}} \right| = b$$
[ length of perpendicular from $$\left( {a,\,0} \right)$$  to (1) $$=$$ radius $$b$$ ]
$$\eqalign{ & \Rightarrow am - b\sqrt {1 + {m^2}} = \pm b\sqrt {1 + {m^2}} \cr & \Rightarrow am - 2b\sqrt {1 + {m^2}} = 0 \cr & {\text{or, }}am = 0\,\,\left( {{\text{not possible as }}a,\,m > 0} \right) \cr & \Rightarrow {a^2}{m^2} = 4{b^2}\left( {1 + {m^2}} \right) \cr & \Rightarrow {m^2} = \frac{{4{b^2}}}{{{a^2} - 4{b^2}}} \cr & \Rightarrow m = \frac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}\,\,\,\left( {\because m > 0} \right) \cr} $$

Given, $$x = 2 - 3\,\sec \,t,\,y = 1 + 4\,\tan \,t$$
$$ \Rightarrow \sec \,t = \frac{{x - 2}}{{ - 3}},\,\tan \,t = \frac{{y - 1}}{4}$$
Since, $${\sec ^2}t - {\tan ^2}t = 1$$
$$\therefore \,\frac{{{{\left( {x - 2} \right)}^2}}}{9} - \frac{{{{\left( {y - 1} \right)}^2}}}{{16}} = 1$$
which is a hyperbola with centre at $$\left( {2,\,1} \right)$$  and eccentricity $$e,$$ given by
$$\eqalign{ & 16 = 9\left( {{e^2} - 1} \right) \cr & \Rightarrow 9{e^2} = 25 \cr & \Rightarrow {e^2} = \frac{{25}}{9} \cr & \Rightarrow e = \frac{5}{3} \cr} $$

$$P = \left( {1,\,0} \right)\,\,Q = \left( {h,\,k} \right)$$     Such that $${K^2} = 8h$$
Let $$\left( {\alpha ,\,\beta } \right)$$  be the midpoint of $$PQ$$
$$\eqalign{ & \alpha = \frac{{h + 1}}{2},\,\,\,\,\,\,\,\,\,\,\beta = \frac{{k + 0}}{2}\, \cr & 2\alpha - 1 = h\,\,\,\,\,\,\,\,\,\,2\beta = k \cr & {\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right)\,\, \Rightarrow {\beta ^2} = 4\alpha - 2 \cr & \Rightarrow {y^2} - 4x + 2 = 0 \cr} $$

$${x^2} - 5xy + 6{y^2} = 0$$     represents a pair of straight lines given by $$x-3y=0$$   and $$x-2y=0.$$
Also $$a{x^2} + b{y^2} + c = 0$$    will represent a circle if $$a=b$$  and $$c$$ is of sign opposite to that of $$a.$$

Let $$P\left( {x,\,y} \right)$$  be the point dividing the join of $$A$$ and $$B$$ in the ratio $$2 : 3$$  internally, then
$$\eqalign{ & x = \frac{{20\,\cos \,\theta + 15}}{5} = 4\,\cos \,\theta + 3 \cr & \Rightarrow \cos \,\theta = \frac{{x - 3}}{4}......\left( {\text{i}} \right) \cr & y = \frac{{20\,\sin \,\theta + 0}}{5} = 4\,\sin \,\theta \cr & \Rightarrow \sin \,\theta = \frac{y}{4}......\left( {{\text{ii}}} \right) \cr} $$
Squaring and adding $$\left( {{\text{i}}} \right)$$ and $$\left( {{\text{ii}}} \right),$$ we get the required locus $${\left( {x - 3} \right)^2} + {y^2} = 16,$$    which is a circle.

Given parabola is $$y = \frac{{{a^3}{x^2}}}{3} + \frac{{{a^2}x}}{2} - 2a$$
$$\eqalign{ & \Rightarrow y = \frac{{{a^3}}}{3}\left( {{x^3} + \frac{3}{{2a}}x + \frac{9}{{16{a^2}}}} \right) - \frac{{3a}}{{16}} - 2a \cr & \Rightarrow y + \frac{{35a}}{{16}} = \frac{{{a^3}}}{3}{\left( {x + \frac{3}{{4a}}} \right)^2} \cr} $$
$$\therefore $$ Vertex of parabola is $$\left( {\frac{{ - 3}}{{4a}},\,\frac{{ - 35a}}{{16}}} \right)$$
To find locus of this vertex,
$$\eqalign{ & x = \frac{{ - 3}}{{4a}}{\text{ and }}y = \frac{{ - 35a}}{{16}} \cr & \Rightarrow a = \frac{{ - 3}}{{4x}}{\text{ and }}a = - \frac{{16y}}{{35}} \cr & \Rightarrow \frac{{ - 3}}{{4x}} = \frac{{ - 16y}}{{35}} \cr & \Rightarrow 64xy = 105 \cr & \Rightarrow xy = \frac{{105}}{{64}}{\text{ which is the required locus}}{\text{.}} \cr} $$

Given equation of ellipse can be written as
$$\frac{{{x^2}}}{6} + \frac{{{y^2}}}{2} = 1\,\,\,\,\,\,\, \Rightarrow {a^2} = 6,\,\,\,{b^2} = 2$$
Now, equation of any variable tangent is
$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} .....({\text{i}})$$
where $$m$$ is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
$$y = \frac{{ - x}}{m}.....({\text{ii}})$$
Eliminating $$m,$$  we get
$$\eqalign{ & \left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2} \cr & \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2} \cr & \Rightarrow \boxed{{{\left( {{x^2} + {y^2}} \right)}^2} = 6{x^2} + 2{y^2}} \cr} $$

Given curve is
$${x^2} + 2xy - 3{y^2} = 0.....(1)$$
Differentiate w.r.t. $$x,\,\,\,2x + 2x\frac{{dy}}{{dx}} + 2y - 6y\frac{{dy}}{{dx}} = 0$$
$${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,\,1} \right)}} = 1$$
Equation of normal at $$\left( {1,\,1} \right)$$  is
$$y = 2 - x.....(2)$$
Solving equation (1) and (2), we get $$x=1 ,\,3$$
Point of intersection $$\left( {1,\,1} \right),\,\left( {3,\, - 1} \right)$$
Normal cuts the curve again in 4th quadrant.

The triangle is formed by the lines

$$\eqalign{ & AB:\left( {1 + p} \right)x - py + p\left( {1 + p} \right) = 0 \cr & AC:\left( {1 + q} \right)x - qy + q\left( {1 + q} \right) = 0 \cr & BC:y = 0 \cr} $$
So that the vertices are
$$A\left( {pq,\left( {p + 1} \right)\left( {q + 1} \right)} \right),\,\,B\left( { - p,\,0} \right),\,\,C\left( { - q,\,0} \right)$$
Let $$H\left( {h,\,k} \right)$$  be the orthocenter of $$\Delta ABC.$$   Then as $$AH \bot BC$$   and passes through $$A\left( {pq,\left( {p + 1} \right)\left( {q + 1} \right)\,} \right)$$
The equation of $$AH$$  is $$x=pq$$
$$\therefore h = pq.....(1)$$
Also $$BH$$  is perpendicular to $$AC$$
$$\eqalign{ & \therefore {m_1}{m_2} = - 1 \Rightarrow \frac{{k - 0}}{{h + p}} \times \frac{{1 + q}}{q} = - 1 \cr & \Rightarrow \frac{k}{{pq + p}} \times \frac{{1 + q}}{q} = - 1\,\,\,\,\,\,\left( {{\text{using equation (1)}}} \right) \cr & \Rightarrow k = - pq.....(2) \cr} $$
From (1) and (2) we observe $$h+k=0$$
$$\therefore $$ Locus of $$\left( {h,\,k} \right)$$  is $$x+y=0$$  which is a straight line.

$$\eqalign{ & x = 2\left( {\cos \,t + \sin \,t} \right) \cr & y\, = 5\left( {\cos \,t - \sin \,t} \right) \cr & \Rightarrow \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 2 \Rightarrow {\text{Ellipes}} \cr} $$