Locus MCQ Questions & Answers in Geometry | Maths

If the tangent $$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $$     passes through $$\left( {{x_1},\,{y_1}} \right)$$  then
$$\eqalign{ & {\left( {{y_1} - m{x_1}} \right)^2} = {a^2}{m^2} + {b^2} \cr & {\text{or, }}\left( {x_1^2 - {a^2}} \right){m^2} - 2{x_1}{y_1} m + y_1^2 - {b^2} = 0 \cr} $$
If the roots be $${m_1}$$ and $${m_2}$$ then
$${m_1} + {m_2} = \frac{{2{x_1}{y_1}}}{{x_1^2 - {a^2}}}{\text{ }}\,{\text{and }}{m_1}{m_1} = \frac{{y_1^2 - {b^2}}}{{x_1^2 - {a^2}}}$$
Given : $${\tan ^2}{\theta _1} + {\tan ^2}{\theta _2} = {\text{constant}} = k\,\left( {{\text{say}}} \right)$$
$$\eqalign{ & \Rightarrow m_1^2 + m_2^2 = k \cr & {\text{or }}{\left( {{m_1} + {m_2}} \right)^2} - 2{m_1}{m_2} = k \cr & \Rightarrow \frac{{4{x^2}{y^2}}}{{{{\left( {x_1^2 - {a^2}} \right)}^2}}} - \frac{{2\left( {y_1^2 - {b^2}} \right)}}{{{{\left( {x_1^2 - {a^2}} \right)}^2}}} = k \cr & {\text{or }}4x_1^2y_1^2 - 2\left( {x_1^2 - {a^2}} \right)\left( {y_1^2 - {b^2}} \right) = k{\left( {x_1^2 - {a^2}} \right)^2} \cr} $$
Hence locus of $$P$$ is $$4{x^2}{y^2} - 2\left( {{x^2} - {a^2}} \right)\left( {{y^2} - {b^2}} \right) = k{\left( {{x^2} - {a^2}} \right)^2}$$

Given that $$\frac{{{x^2}}}{{1 - r}} - \frac{{{y^2}}}{{1 + r}} = 1,\,\,\,r > 1$$
$$\eqalign{ & {\text{As }}r > 1 \cr & \therefore 1 - r < 0{\text{ and }}1 + r > 0 \cr & \therefore {\text{Let }}1 - r = - {a^2},\,\,\,1 + r = {b^2},\,\,{\text{then we get}} \cr & \frac{{{x^2}}}{{ - {a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,\,\,\,\,\,\, \Rightarrow \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = - 1 \cr} $$
which is not possible for any real values of $$x$$ and $$y.$$

Tangent to the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$   is $$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $$
Given that $$y = \alpha x + \beta $$   is the tangent of hyperbola
$$\eqalign{ & \Rightarrow m = \alpha {\text{ and }}{a^2}{m^2} - {b^2} = {\beta ^2} \cr & \therefore {a^2}{\alpha ^2} - {b^2} = {\beta ^2} \cr} $$
Locus is $${a^2}{x^2} - {y^2} = {b^2}$$    which is hyperbola.

If $$\left( {h,\,k} \right)$$  is the mid point of line joining focus $$\left( {a,\,0} \right)$$  and $$Q\left( {a{t^2},\,2at} \right)$$   on parabola then $$h = \frac{{a + a{t^2}}}{2},\,k = at$$
Eliminating $$t,$$  we get $$2h = a + a\left( {\frac{{{k^2}}}{{{a^2}}}} \right)$$
$$ \Rightarrow {k^2} = a\left( {2h - a} \right)\,\,\,\, \Rightarrow {k^2} = 2a\left( {h - \frac{a}{2}} \right)$$
$$\therefore $$ Locus of $$\left( {h,\,k} \right)$$  is $${y^2} = 2a\left( {x - \frac{a}{2}} \right)$$
whose directrix is $$\left( {x - \frac{a}{2}} \right) = - \frac{a}{2}$$
$$ \Rightarrow x = 0$$

We have $$2{x^2} + 3{y^2} - 8x - 18y + 35 = k$$
$$ \Rightarrow 2{\left( {x - 2} \right)^2} + 3{\left( {y - 3} \right)^2} = k$$
For $$k=0,$$   we get $$ \Rightarrow 2{\left( {x - 2} \right)^2} + 3{\left( {y - 3} \right)^2} = 0$$      which represents the point $$\left( {2,\,3} \right).$$

Let the given line be the $$y$$-axis and the circle to have the equation $${x^2} + {y^2} + 2gx + 2fy + c = 0$$
then according to given condition
$$\eqalign{ & {x^2} = {\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} - \left( {{g^2} + {f^2} - c} \right) \cr & \Rightarrow {\left( {y + f} \right)^2} = - 2g\left( {x - \frac{{{f^2} - c}}{{2g}}} \right), \cr} $$
which represents a parabola with its axis $$ \bot $$ to $$y$$-axis.

$$A = \left\{ {\left( {a,\,b} \right) \in R \times R:\left| {a - 5} \right| < 1,{\text{ }}\left| {b - 5} \right| < 1} \right\}$$
Let $$a-5=x, \,\,b-5=y$$
Set $$A$$ contains all points inside $$\left| x \right| < 1,\,\left| y \right| < 1$$
$$B = \left\{ {\left( {a,\,b} \right) \in R \times R:4{{\left( {a - 6} \right)}^2} + 9{{\left( {b - 5} \right)}^2} \leqslant 36} \right\}$$
Set $$B$$ contains all points inside or on
$$\frac{{{{\left( {x - 1} \right)}^2}}}{9} + \frac{{{y^2}}}{4} = 1$$

$$\therefore \left( { \pm 1,\, \pm 1} \right)$$   lies inside the ellipse.
Hence, $$A \subset B$$