Parabola MCQ Questions & Answers in Geometry | Maths

Equation of tangent at $$P\left( {16,\,16} \right)$$   is given as :
$$x - 2y + 16 = 0$$

Slope of $$PC\left( {{m_1}} \right) = \frac{4}{3}$$
Slope of $$PB\left( {{m_2}} \right) = - 2$$
Hence, $$\tan \,\theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}.{m_2}}}} \right| = \left| {\frac{{\frac{4}{3} + 2}}{{1 - \frac{4}{3}.2}}} \right|$$
$$ \Rightarrow \tan \,\theta = 2$$

Any tangent to the parabola $${y^2} = 8ax$$   is
$$y = mx + \frac{{2a}}{m}.....({\text{i}})$$
If (i) is a tangent to the circle, $${x^2} + {y^2} = 2{a^2}$$   then,
$$\eqalign{ & \sqrt 2 a = \pm \frac{{2a}}{{m\sqrt {{m^2} + 1} }} \cr & \Rightarrow {m^2}\left( {1 + {m^2}} \right) = 2 \cr & \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0 \cr & \Rightarrow m = \pm 1 \cr} $$
So from (i), $$y = \pm \left( {x + 2a} \right)$$

The vertex is the middle point of the perpendicular dropped from the focus to the directrix.

$$y$$-coordinate of the centroid $$ = \frac{{{y_1} + {y_2} + {y_3}}}{3} = \frac{0}{3} = 0.$$

Let common tangent be $$y = mx + \frac{{\sqrt 5 }}{m}$$
Since, perpendicular distance from centre of the circle to the common tangent is equal to radius of the circle, therefore $$\frac{{\frac{{\sqrt 5 }}{m}}}{{\sqrt {1 + {m^2}} }} = \sqrt {\frac{5}{2}} $$
On squaring both the side, we get
$$\eqalign{ & {m^2}\left( {1 + {m^2}} \right) = 2 \cr & \Rightarrow {m^4} + {m^2} - 2 = 0 \cr & \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0 \cr & \Rightarrow m = \pm 1\,\,\,\left( {\because m \ne \pm \sqrt 2 } \right) \cr} $$
$$y = \pm \left( {x + \sqrt 5 } \right),$$    both statements are correct as $$m = \pm 1$$
satisfies the given equation of statement-2.

Let the two tangents to the parabola $${y^2} = 4ax$$   be $$PT$$  and $$QT$$  which are at right angle to one another at $$T\left( {h,\,k} \right).$$   Then we have to find the locus of $$T\left( {h,\,k} \right).$$
We know that $$y = mx + \frac{a}{m},$$    where $$m$$ is the slope is the equation of tangent to the parabola $${y^2} = 4ax$$   for all $$m.$$

Since this tangent to the parabola will pass through $$T\left( {h,\,k} \right)$$  so $$k = mh + \frac{a}{m}\,;\,\,{\text{or}}\,{\text{ }}{m^2}h - mk + a = 0$$
This is a quadratic equation in $$m$$ so will have two roots, say $${m_1}$$ and $${m_2},$$ then
$${m_1} + {m_2} = \frac{k}{h},\,\,{\text{and}}\,{\text{ }}{m_1}:{m_2} = \frac{a}{h}$$
Given that the two tangents intersect at right angle so $${m_1}.{m_2} = - 1{\text{ or }}\frac{a}{h} = - 1{\text{ or }}h + a = 0$$
The locus of $$T\left( {h,\,k} \right)$$  is $$x + a = 0,$$   which is the equation of directrix.

$$\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$$        though $${h^2} = ab$$   is satisfied. So the equation represents a pair of lines.

If the middle point of a chord is $$\left( {\alpha ,\,\beta } \right)$$  then $$\alpha = \frac{{2{t^2} + 0}}{2},\,\,\beta = \frac{{4t + 0}}{2}$$
Eliminating $$t,\,\alpha = {\left( {\frac{\beta }{2}} \right)^2}.$$   So, the locus is $${y^2} = 4x.$$

Given parabola is $${y^2} = 4x......\left( 1 \right)$$
Let $$P \equiv \left( {t_1^2,\,2{t_1}} \right){\text{ and }}Q \equiv \left( {t_2^2,\,2{t_2}} \right)$$
Slope of $$OP = \frac{{2{t_1}}}{{t_1^2}} = \frac{2}{{{t_1}}}$$     and slope of $$OQ = \frac{2}{{{t_2}}}$$
Since $$OP \bot OQ,$$
$$\therefore \,\frac{4}{{{t_1}{t_2}}} = - 1{\text{ or }}{t_1}{t_2} = - 4......\left( 2 \right)$$
Let $$R\left( {h,\,k} \right)$$  be the middle point of $$PQ,$$  then
$$h = \frac{{t_1^2 + t_2^2}}{2}......\left( 3 \right)$$
and $$k = {t_1} + {t_2}......\left( 4 \right)$$
From $$\left( 4 \right),\,{k^2} = t_1^2 + t_2^2 + 2{t_1}{t_2} = 2h - 8\,\,\,\,\left[ {{\text{From}}\left( 2 \right){\text{and}}\left( 3 \right)} \right]$$
Hence locus of $$R\left( {h,\,k} \right)$$  is $${y^2} = 2x - 8.$$

The length of the perpendicular drawn from the given focus upon the given line $$x - y + 1 = 0$$    is $$\frac{{0 - 0 + 1}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}} }} = \frac{1}{{\sqrt 2 }}$$
The directrix is parallel to the tangent at the vertex.
So, the equation of the directrix is $$x - y + \lambda = 0,$$    where $$\lambda $$ is a constant to be determine.
But the distance between the focus and the directrix $$ = 2 \times $$  (the distance between the focus and the tangent at the vertex)
$$ = 2 \times \frac{1}{{\sqrt 2 }} = \sqrt 2 $$
Hence $$\frac{{0 - 0 + \lambda }}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}} }} = \sqrt 2 $$
$$\therefore \,\lambda = 2.$$   [$$\lambda $$ must be positive see figure]
$$\therefore $$  The directrix is the line $$x - y + 2 = 0.$$
Let $$\left( {x,\,y} \right)$$  be a moving point on the parabola. By the focus-directrix property of the parabola, its equation is
$$\eqalign{ & {\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {\left( { \pm \frac{{x - y + 2}}{{\sqrt 2 }}} \right)^2} \cr & {\text{or, }}\,{x^2} + {y^2} + 2xy - 4x + 4y - 4 = 0 \cr} $$