Parabola MCQ Questions & Answers in Geometry | Maths

$$R = \left( {1,\,\frac{{1 + 3\lambda }}{{1 + \lambda }}} \right).$$    It is an interior point of $${y^2} - 4x = 0$$   if $${\left( {\frac{{1 + 3\lambda }}{{1 + \lambda }}} \right)^2} - 4 < 0$$
Therefore, $$ - \frac{3}{5} < \lambda < 1.$$   But $$\lambda > 0.$$

Let tangent to $${y^2} = 4x$$   be $$y = mx + \frac{1}{m}$$
Since this is also tangent to $${x^2} = - 32y$$
$$\therefore {x^2} = - 32\left( {mx + \frac{1}{m}} \right)\,\,\, \Rightarrow {x^2} + 32mx + \frac{{32}}{m} = 0$$
Now, $$D=0$$
$$\eqalign{ & {\left( {32} \right)^2} - 4\left( {\frac{{32}}{m}} \right) = 0 \cr & \Rightarrow {m^3} = \frac{4}{{32}} \cr & \Rightarrow {m^3} = \frac{1}{8} \cr & \Rightarrow m = \frac{1}{2} \cr} $$

$${y^2} - 2y + 1 = x - 1{\text{ or }}{\left( {y - 1} \right)^2} = x - 1$$
Putting $$x - 1 = X,\,\,y - 1 = Y$$     the equation becomes $${Y^2} = X,$$   i.e., $${Y^2} = 4.\frac{1}{4}.X.$$    So, the focus $$ = {\left( {\frac{1}{4},\,0} \right)_{X,\,Y}} = \left( {\frac{5}{4},\,1} \right).$$

Minimum distance $$ \Rightarrow $$ perpendicular distance
Equation of normal at $$p\left( {2{t^2},\,4t} \right)$$
$$y = - tx + 4t + 2{t^3}$$
It passes through $$C\left( {0,\, - 6} \right)\,\, \Rightarrow {t^3} + 2t + 3 = 0\,\, \Rightarrow t = - 1$$

Centre of new circle $$ = P\left( {2{t^2},\,4t} \right) = P\left( {2,\, - 4} \right)$$
Radius $$ = PC = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} = 2\sqrt 2 $$
$$\therefore $$ Equation of the circle is
$$\eqalign{ & {\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2} \cr & \Rightarrow {x^2} + {y^2} - 4x + 8y + 12 = 0 \cr} $$

Here, $${y^2} - 2y + 1 = 4\left( {x + 1} \right){\text{ or }}{\left( {y - 1} \right)^2} = 4\left( {x + 1} \right)$$
So, the axis is $$y - 1 = 0.$$   Also $$\left( { - 2,\,1} \right)$$  lies on the axis, and it is exterior to the parabola because $${1^2} - 4\left( { - 2} \right) - 2\left( 1 \right) - 3 > 0.$$      Hence, only one normal is possible.

We know the tangent to the parabola $${y^2} = 4ax$$   at $$\left( {a{t^2},\,2at} \right)$$   is $$ty = x + a{t^2}.$$
Here $$a = 2$$
So, the tangent at $$\left( {2{t^2},\,4t} \right)$$   to the parabola
$${y^2} = 8x$$   is $$ty = x + 2{t^2}......\left( {\text{i}} \right)$$
$$'m'$$ of $$\left( {\text{i}} \right)$$ is $$\frac{1}{t};\left( {\text{i}} \right)$$  makes $${45^ \circ }$$  with $$y = 3x + 5$$   if
$$\eqalign{ & \tan \,{45^ \circ } = \left| {\frac{{\frac{1}{t} - 3}}{{1 + \frac{1}{t}.3}}} \right| = \left| {\frac{{1 - 3t}}{{t + 3}}} \right| \cr & \therefore \,1 = \left| {\frac{{1 - 3t}}{{t + 3}}} \right|{\text{; or }}\frac{{1 - 3t}}{{t + 3}} = \pm 1\,;{\text{ or }}1 - 3t = t + 3,\, - t - 3 \cr & \therefore \,4t = - 2{\text{ or }}2t = 4 \cr & \therefore \,t = - \frac{1}{2}{\text{ or }}2 \cr} $$
Putting in $$\left( {\text{i}} \right),$$ the tangents have the equations
$$\eqalign{ & - \frac{1}{2}y = x + 2.\frac{1}{4}{\text{ i}}{\text{.e}}{\text{., }}2x + y + 1 = 0 \cr & {\text{and }}2y = x + 2.4{\text{ i}}{\text{.e}}{\text{., }}x - 2y + 8 = 0 \cr} $$

Any tangent to $${y^2} = 4ax$$   is $$y = mx + \frac{a}{m}.$$   It touches $${x^2} = 4ay$$   if $${x^2} = 4a\left( {mx + \frac{a}{m}} \right),$$     i.e., $${x^2} - 4amx - \frac{{4{a^2}}}{m} = 0$$     has equal roots.
So, $${m^3} + 1 = 0,$$   i.e., $$m = - 1.$$   Hence, the common tangent is $$y = - x - a.$$

Equation of tangent at $$\left( {1,\,7} \right)$$  to $${x^2} = y - 6$$   is $$2x - y + 5 = 0.$$

Now, perpendicular from centre $$O\left( { - 8,\, - 6} \right)$$   to $$2x-y+5=0$$    should be equal to radius of the circle
$$\eqalign{ & \therefore \left| {\frac{{ - 16 + 6 + 5}}{{\sqrt 5 }}} \right| = \sqrt {64 + 36 - C} \cr & \Rightarrow \sqrt 5 = \sqrt {100 - c} \cr & \Rightarrow c = 95 \cr} $$

Vertex $$ = \left( {0,\,0} \right)$$   The ends of latus rectum are $$\left( {2,\,4} \right)\left( {2,\, - 4} \right)$$
$$\therefore $$  centre $$ = \left( {0,\,0} \right)$$   radius $$ = \sqrt {{2^2} + {4^2}} = \sqrt {20} $$

$$9{y^2} - 12y + 4 = 16x + 61\,\,{\text{or }}{\left( {3y - 2} \right)^2} = 4.4\left( {x + \frac{{61}}{{16}}} \right)$$
Therefore, the equation of the axis is $$3y - 2 = 0.$$