Parabola MCQ Questions & Answers in Geometry | Maths

$$\eqalign{ & \frac{{dy}}{{dx}} = 2x - 5 \cr & \therefore {m_1} = {\left( {2x - 5} \right)_{\left( {2,\,0} \right)}} = - 1, \cr & {m_2} = {\left( {2x - 5} \right)_{\left( {3,\,0} \right)}} = 1 \cr & \Rightarrow {m_1}{m_2} = - 1, \cr} $$
i.e., the tangents are perpendicular to each other.

Let $$A\left( {x,\,y} \right) = \left( {{t^2},\,2t} \right)$$     be any point on parabola $${y^2} = 4x.$$

Let $$P\left( {h,\,k} \right)$$  divides $$OA$$  in theratio 1 : 3
Then $$\left( {h,\,k} \right) = \left( {\frac{{{t^2}}}{4},\,\frac{{2t}}{4}} \right)$$
$$ \Rightarrow h = \frac{{{t^2}}}{4}$$   and $$k = \frac{t}{2}\,\, \Rightarrow h = {k^2}$$
$$\therefore $$ locus of $$P\left( {h,\,k} \right)$$   is $$x = {y^2}$$

The equation of the axis of the parabola is $$y - 2 = 0,$$   which is parallel to the $$x$$-axis. So, a ray parallel to the $$x$$-axis is parallel to the axis of the parabola. We know that any ray parallel to the axis of a parabola passes through the focus after reflection. Here $$\left( {0,\,2} \right)$$  is the focus.

Solving equations of parabolas
$${y^2} = 4ax$$   and $${x^2} = 4ay$$
we get $$\left( {0,\,0} \right)$$  and $$\left( {4a,\,4a} \right)$$
Substituting in the given equation of line
$$2bx + 3cy + 4d = 0,$$
we get $$d=0$$  and $$2b + 3c = 0$$
$$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$

For parabola $${y^2} = 16x,$$   focus $$ \equiv \left( {4,\,0} \right).$$
Let $$m$$ be the slope of focal chord then equation is
$$y = m\left( {x - 4} \right).....(1)$$
But given that above is a tangent to the circle
$${\left( {x - 6} \right)^2} + {y^2} = 2$$
With Centre, $$C\left( {6,\,0} \right),\,r = \sqrt 2 $$
$$\therefore $$ Length of $${ \bot ^{{\text{lar}}}}$$  from (6, 0) to (1) $$=r$$
$$\eqalign{ & \Rightarrow \frac{{6m - 4m}}{{\sqrt {{m^2} + 1} }} = \sqrt 2 \cr & \Rightarrow 2m = \sqrt {2\left( {{m^2} + 1} \right)} \cr & \Rightarrow 2{m^2} = {m^2} + 1 \cr & \Rightarrow {m^2} = 1 \cr & \Rightarrow m = \pm 1 \cr} $$

Let the tangent to $${y^2} = 8x$$   be $$y = mx + \frac{2}{m}$$
If it is common tangent to parabola and circle, then $$y = mx + \frac{2}{m}$$   is a tangent to $${x^2} + {y^2} = 2$$
$$\eqalign{ & \therefore \left| {\frac{{\frac{2}{m}}}{{\sqrt {{m^2} + 1} }}} \right| = \sqrt 2 \cr & \Rightarrow \frac{4}{{{m^2}\left( {1 + {m^2}} \right)}} = 2 \cr & \Rightarrow {m^4} + {m^2} - 2 = 0 \cr & \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0 \cr & \Rightarrow m = 1\,\,\,{\text{or }} - 1 \cr} $$
$$\therefore $$ Required tangents are $$y=x+2$$   and $$y=-x-2$$
Their common point is $$\left( { - 2,\,0} \right)$$
$$\therefore $$ Tangents are drawn from $$\left( { - 2,\,0} \right)$$
$$\therefore $$ Chord of contact $$PQ$$  to circle is
$$x.\left( { - 2} \right) + y.0 = 2{\text{ or }}x = - 1$$
and Chord of contact $$RS$$  to parabola is
$$y.0 = 4\left( {x - 2} \right){\text{ or }}x = 2$$
Hence coordinates of $$P$$ and $$Q$$ are $$\left( { - 1,\,1} \right){\text{ and }}\left( { - 1,\, - 1} \right)$$
Also coordinates of $$R$$ and $$S$$ are $$\left( {2,\, - 4} \right){\text{ and }}\left( {2,\,4} \right)$$
$$\therefore $$ Area of trapezium $$PQRS$$   is $$\frac{1}{2}\left( {2 + 8} \right) \times 3 = 15$$

$$\eqalign{ & x = a\left( {{y^2} + \frac{b}{a}y + \frac{{{b^2}}}{{4{a^2}}}} \right) - \frac{{{b^2}}}{{4a}} + c \cr & {\text{or }}a{\left( {y + \frac{b}{{2a}}} \right)^2} = x + \frac{{{b^2}}}{{4a}} - c \cr & {\text{or }}{\left( {y + \frac{b}{{2a}}} \right)^2} = \frac{1}{a}\left\{ {x + \frac{{{b^2} - 4ac}}{{4a}}} \right\} \cr} $$
$$\therefore $$  the latus rectum $$ = \frac{1}{a}.$$

Any normal to $${y^2} = 4x$$  at $$\left( {{t^2},\,2t} \right)$$  is $$y + tx = 2t + {t^3}.$$    Comparing this with $$y = x + k,$$   we get $$\frac{t}{{ - 1}} = \frac{{2t + {t^3}}}{k}{\text{ or }}2 + {t^2} = - k$$
$$\therefore \,\,k$$  is negative.

Here the lines are two tangents to the parabola $${y^2} = 4ax$$   passing through the point $$\left( { - a,\, - b} \right),$$   their slopes being $${m_1}$$ and $${m_2}.$$  But $$\left( { - a,\, - b} \right)$$   is on the directrix $$x + a = 0.$$   So, the tangents are at right angles.

$$\eqalign{ & {y^2} + 4y + 4x + 2 = 0 \cr & {y^2} + 4y + 4 = - 4x + 2 \cr & {\left( {y + 2} \right)^2} = - 4\left( {x - \frac{1}{2}} \right) \cr} $$
It is of the form $${Y^2} = - 4AX$$
whose directrix is given by $$X=A$$
$$\therefore $$ Required equation is $$x - \frac{1}{2} = 1\,\,\, \Rightarrow x = \frac{3}{2}$$