Parabola MCQ Questions & Answers in Geometry | Maths

$${y^2} + 4 = 4x{\text{ or }}{y^2} = 4\left( {x - 1} \right)$$
Any tangent to it is $$y = m\left( {x - 1} \right) + \frac{1}{m}.$$     It is through the origin if $$0 = - m + \frac{1}{m}\,\,\,\, \Rightarrow m = 1,\, - 1$$
So, the two tangents are inclined at $$\frac{\pi }{2}.$$

Here, $$\alpha = \frac{{at_1^2 + at_2^2}}{2},\,\,\beta = \frac{{2a{t_1} + 2a{t_2}}}{2} = a\left( {{t_1} + {t_2}} \right)$$
Also, $$\frac{{2a{t_1} - 0}}{{at_1^2 - 0}} \times \frac{{2a{t_2} - 0}}{{at_2^2 - 0}} = - 1\,\,\,\,\,\,\,\, \Rightarrow {t_1}{t_2} = - 4$$
Now, $$\frac{{2\alpha }}{a} = t_1^2 + t_2^2 = {\left( {{t_1} + {t_2}} \right)^2} - 2{t_1}{t_2} = {\left( {\frac{\beta }{\alpha }} \right)^2} - 2.\left( { - 4} \right)$$
$$\therefore $$  the locus of $$\left( {\alpha ,\,\beta } \right)$$  is $$\frac{{2x}}{a} = \frac{{{y^2}}}{{{a^2}}} + 8,$$    which is a parabola.

Any tangent to the parabola $${y^2} = 4x$$  is $$y = mx + \frac{1}{m}.$$   It passes through $$\left( {\alpha ,\,\beta } \right)$$  if $$\beta = m\alpha + \frac{1}{m}{\text{ or }}\alpha {m^2} - \beta m + 1 = 0$$
It will have roots $${m_1},\,2{m_1}$$  if $${m_1} + 2{m_1} = \frac{\beta }{\alpha },\,\,{m_1}.2{m_1} = \frac{1}{\alpha }\,\,\,\,\,\, \Rightarrow 2.{\left( {\frac{\beta }{{3\alpha }}} \right)^2} = \frac{1}{\alpha }$$

$$\Delta \ne 0{\text{ for all real }}\lambda {\text{ and }}{h^2} = ab\,\,\,\,\, \Rightarrow {2^2} = \lambda .1$$

Here, $${\left( {x - 1} \right)^2} + {\left( {y - 3} \right)^2} = {\left\{ {\frac{{5x - 12y + 17}}{{\sqrt {{5^2} + {{\left( { - 12} \right)}^2}} }}} \right\}^2}$$
$$\therefore $$  the focus $$ = \left( {1,\,3} \right)$$  and the directrix is $$5x - 12y + 17 = 0$$
The distance of the focus from the directrix $$ = \left| {\frac{{5 \times 1 - 12 \times 3 + 17}}{{\sqrt {{5^2} + {{\left( { - 12} \right)}^2}} }}} \right| = \frac{{14}}{{13}}$$
$$\therefore $$  latus rectum $$ = 2 \times \frac{{14}}{{13}} = \frac{{28}}{{13}}.$$

Parabola $${y^2} = 8x$$

We know that the locus of point of intersection of two perpendicular tangents to a parabola is its directrix.
Point must be on the directrix of parabola
$$\because $$ equation of directrix $$x + 2 = 0\,\,\, \Rightarrow x = - 2$$
Hence the point is $$\left( { - 2,\,0} \right)$$

Equation of the normal to a parabola $${y^2} = 4bx$$   at point $$\left( {bt_1^2,\,2b{t_1}} \right)$$   is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$
As given, it also passes through $$\left( {bt_2^2,\,2b{t_2}} \right)$$  then
$$\eqalign{ & 2b{t_2} = - {t_1}bt_2^2 + 2b{t_1} + bt_1^3 \cr & 2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right) \cr & \Rightarrow 2{t_2} - 2{t_1} = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right) \cr & \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right) \cr & \Rightarrow {t_2} + {t_1} = - \frac{2}{{{t_1}}} \cr & \Rightarrow {t_2} = - {t_1} - \frac{2}{{{t_1}}} \cr} $$

$$L,\,L'$$  are the ends of the latus rectum. $$S$$ bisects $$LL'.\,VSV'$$   is the perpendicular bisector of $$LL',$$   where $$VS = \frac{1}{4}LL' = V'S.$$     Clearly, two parabolas are possible.

The focus $$ = \left( {\frac{{3 + 3}}{2},\,\frac{{2 - 2}}{2}} \right) = \left( {3,\,0} \right).$$     The vertex $$ = \left( {2,\,0} \right)$$
As $$MV = VS,\,\,M = \left( {1,\,0} \right).$$     Clearly, the directrix is perpendicular to $$VS$$  whose equation is $$y = 0.$$  So, the directrix is $$x = k$$  which passes through $$M\left( {1,\,0} \right).$$   Therefore, we get $$x = 1.$$
$$\therefore $$  the equation of the parabola is $${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} = {\left( {\frac{{x - 1}}{{\sqrt 1 }}} \right)^2}.$$

Let $$P = \left( {\alpha ,\,\beta } \right).$$   Any tangent to the parabola is $$y = mx + \frac{a}{m}.$$   It passes through $$\left( {\alpha ,\,\beta } \right).$$  So, $$\beta = m\alpha + \frac{1}{m}\left( {\because {\text{ here }}a = 1} \right);\,\,\therefore \,{m^2}\alpha - \beta m + 1 = 0$$
Its roots are $${m_1},\,3{m_1}.$$  So, $${m_1} + 3{m_1} = \frac{\beta }{\alpha },\,\,\,{m_1}.3{m_1} = \frac{1}{\alpha }$$
$$\therefore \,\,3.{\left( {\frac{\beta }{{4\alpha }}} \right)^2} = \frac{1}{\alpha }{\text{ or }}3{\beta ^2} = 16\alpha $$
Thus, the locus is $$3{y^2} = 16x,$$   which is a parabola.