1.
$$AB$$ is a chord of the parabola $${y^2} = 4ax.$$ If its equation is $$y = mx + c$$ and it subtends a right angle at the vertex of the parabola then :
The equation of the pair of lines $$VA$$ and $$VB,$$ where $$V$$ is the vertex $$ = \left( {0,\,0} \right)$$ is $${y^2} = 4ax.\frac{{y - mx}}{c}{\text{ or }}c{y^2} - 4axy + 4am{x^2} = 0$$
The pair is at right angles if $$c + 4am = 0.$$
2.
The equation of the common tangent touching the circle $${\left( {x - 3} \right)^2} + {y^2} = 9$$ and the parabola $${y^2} = 4x$$ above the $$x$$-axis is-
Let the equation of tangent to $${y^2} = 4x$$ be $$y = mx + \frac{1}{m}$$ where $$m$$ is the slope of the tangent.
If it is tangent to the circle $${\left( {x - 3} \right)^2} + {y^2} = 9$$ then length of perpendicular to tangent from centre $$\left( {3,\,0} \right)$$ should be equal to the radius $$3.$$
$$\eqalign{
& \therefore \frac{{3m + \frac{1}{m}}}{{\sqrt {{m^2} + 1} }} = 3 \cr
& \Rightarrow 9{m^2} + \frac{1}{{{m^2}}} + 6 = 9{m^2} + 9 \cr
& \Rightarrow m = \pm \frac{1}{{\sqrt 3 }} \cr} $$
$$\therefore $$ Tangents are $$x - y\sqrt 3 + 3 = 0$$ and $$x + y\sqrt 3 + 3 = 0$$ out of which $$x - y\sqrt 3 + 3 = 0$$ meets the parabola at $$\left( {3,\,2\sqrt 3 } \right)$$ i.e., above $$x$$-axis.
3.
A double ordinate of the parabola $${y^2} = 8px$$ is of length $$16p.$$ The angle subtended by it at the vertex of the parabola is :
Solving the equations,
$$\eqalign{
& {x^2} + \frac{{3\left( {x + 1} \right)}}{2} + 2x = 0 \cr
& {\text{or, }}{x^2} + \frac{{7x}}{2} + \frac{3}{2} = 0 \cr
& {\text{or, }}2{x^2} + 7x + 3 = 0 \cr
& {\text{or, }}\left( {2x + 1} \right)\left( {x + 3} \right) = 0 \cr
& {\text{or, }}x = - \frac{1}{2},\, - 3 \cr} $$
But $$x = - 3$$ makes $$y$$ imaginary because $$2{y^2} = 3\left( {x + 1} \right)$$
So, $$x = - \frac{1}{2}\,\,\,\,\,\, \Rightarrow y = \pm \frac{{\sqrt 3 }}{2}$$
$$\therefore $$ the length of the chord $$=$$ the distance between $$\left( { - \frac{1}{2},\,\frac{{\sqrt 3 }}{2}\,} \right)$$ and $$\left( { - \frac{1}{2},\, - \frac{{\sqrt 3 }}{2}\,} \right).$$
5.
The equation of a parabola is $${y^2} = 4x.\,P\left( {1,\,3} \right)$$ and $$Q\left( {1,\,1} \right)$$ are two points in the $$x$$-$$y$$ plane. Then, for the parabola :
A.
$$P$$ and $$Q$$ are exterior points
B.
$$P$$ is an interior point while $$Q$$ is an exterior point
C.
$$P$$ and $$Q$$ are interior points
D.
$$P$$ is an exterior point while $$Q$$ is an interior point
Answer :
$$P$$ is an exterior point while $$Q$$ is an interior point
Since, the equation of tangent to parabola $${y^2} = 4x$$ is
$$y = mx + \frac{1}{m}.....(1)$$
The line (a) is also the tangent to circle
$${x^2} + {y^2} - 6x = 0$$
Then centre of circle $$ = \left( {3,\,0} \right)$$
radius of circle $$= 3$$
The perpendicular distance from centre to tangent is equal to the radius of circle
$$\eqalign{
& \therefore \frac{{\left| {3m + \frac{1}{m}} \right|}}{{\sqrt {1 + {m^2}} }} = 3 \cr
& \Rightarrow {\left( {3m + \frac{1}{m}} \right)^2} = 9\left( {1 + {m^2}} \right) \cr
& \Rightarrow m = \pm \frac{1}{{\sqrt 3 }} \cr} $$
Then, from equation (1) : $$y = \pm \frac{1}{{\sqrt 3 }}x \pm \sqrt 3 $$
Hence, $$\sqrt 3 y = x + 3$$ is one of the required common tangent.
7.
If one end of a focal chord of the parabola, $${y^2} = 16x$$ is at $$\left( {1,\,4} \right),$$ then the length of this focal chord is :
$$\because {y^2} = 16x\,\, \Rightarrow a = 4$$
One end of focal of the parabola is at (1, 4)
$$\because \,y$$ co-ordinate of focal chord is $$2\,at$$
$$\eqalign{
& \therefore 2\,at = 4 \cr
& \Rightarrow t = \frac{1}{2} \cr} $$
Hence, the required length of focal chord
$$ = a{\left( {t + \frac{1}{t}} \right)^2} = 4 \times {\left( {2 + \frac{1}{2}} \right)^2} = 25$$
8.
The axis of a parabola is along the line $$y = x$$ and the distances of its vertex and focus from origin are $$\sqrt 2 $$ and $$2\sqrt 2 $$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is -
A.
$${\left( {x + y} \right)^2} = \left( {x - y - 2} \right)$$
B.
$${\left( {x - y} \right)^2} = \left( {x + y - 2} \right)$$
C.
$${\left( {x - y} \right)^2} = 4\left( {x + y - 2} \right)$$
D.
$${\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right)$$
Since, distance of vertex from origin is $$\sqrt 2 $$ and focus is $$2\sqrt 2 $$
$$\therefore $$ Vertex is $$\left( {1,\,1} \right)$$ and focus is $$\left( {2,\,2} \right),$$ directrix $$x+y=0$$
Let the parabola be $${y^2} = 4ax.\,\,Q$$ is the intersection of the lines $$x = 0$$ and $$ty = x + a{t^2},$$ where $$P = \left( {a{t^2},\,2at} \right).$$ Solving these, $$Q = \left( {0,\,at} \right).$$ Also $$S = \left( {a,\,0} \right).$$
$$\eqalign{
& \therefore \,\,S{P^2} = {\left\{ {a\left( {{t^2} - 1} \right)} \right\}^2} + 4{a^2}{t^2} = {a^2}{\left( {{t^2} + 1} \right)^2} \cr
& \,\,\,\,\,\,\,\,S{Q^2} = {a^2} + {a^2}{t^2} = {a^2}\left( {{t^2} + 1} \right){\text{ and }}SV = a \cr
& \therefore \,S{Q^2} = SP.SV\, \cr} $$
10.
If two of the three feet of normals drawn from a point to the parabola $${y^2} = 4x$$ be $$\left( {1,\,2} \right)$$ and $$\left( {1,\, - 2} \right)$$ then the third foot is :