Tracing of the parabola - class-XII

$y^2+3=2(2x+y)$ represents parabola.

Distance of vertex from directrix $y -$ axis is $2$ and we know it
is half the distance of focus from directrix.
$\displaystyle \therefore $ Focus is $\displaystyle \left ( 4,0 \right ).$

Given parabola is $y^2=12x$    ....$(i)$
Here $a=3$
For point $P(x, y), y=6$
This point lie on the parabola 
$\therefore (6)^2=12x\Rightarrow x=3$
Now, focal distance of point $P$ is $x+a=6$

Using the definition of a parabola (distance from focus equals distance to directrix), for focus (3/2, 0) and directrix x = -3/2, the equation is sqrt((x-3/2)^2 + y^2) = |x + 3/2|. Squaring both sides gives (x-3/2)^2 + y^2 = (x+3/2)^2, which simplifies to y^2 = 6x.

Slope of axis$=\frac{{5 - 3}}{{3 - 1}} = \frac{2}{2} = 1$
Slope of directrix$=-1$
equation of tangent at vertex $A$
$\begin{array}{l}Pt(1,3),m =  - 1\ \Rightarrow y - 3 =  - 1\left( {x - 1} \right)\ \Rightarrow x + y - 4 = 0\end{array}$
equation of directrix
$\begin{array}{l}x + y = \lambda \a = SA\ = \sqrt {4 + 4}  = 2\sqrt 2 \end{array}$
$A$ is midpoint of $PS$
$\begin{array}{l}\frac{{n + 3}}{2} = 1,\frac{{k + 5}}{2} = 3\ \Rightarrow n =  - 1,k = 1\end{array}$
$(-1,1)$ lies on directrix
$-1+1=\lambda=0$
equation of diectrix: $L:y+x=0$
$\begin{array}{l}QO = QS\ \Rightarrow \left| {\frac{{l + m}}{{\sqrt 2 }}} \right| = \sqrt {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \ \Rightarrow {\left( {l + m} \right)^2} = 2\left[ {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \right]\ \Rightarrow {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 3} \right)}^2} + {{\left( {y - 5} \right)}^2}} \right]\end{array}$

$\Rightarrow y^2 = 4y - 4x$

Putting $x=1,y=2$

Rewrite the equation as 4(y - 3/2)^2 = -12(x - 5/2), which simplifies to (y - 3/2)^2 = -3(x - 5/2). Comparing with standard form reveals the vertex is at (-5/2, 3/2).

For (y-2)^2 = 20(x+3), the vertex is (-3, 2) and 4a = 20, so a = 5. The focus is (h+a, k) = (-3+5, 2) = (2, 2).

4(y^2 - y) = 6x + 5. 4(y - 1/2)^2 = 6x + 5 + 1 = 6x + 6. (y - 1/2)^2 = 3/2(x + 1). Vertex (-1, 1/2), 4a = 3/2, a = 3/8. Directrix is x = h - a = -1 - 3/8 = -11/8. So 8x + 11 = 0.

Rewrite x^2 - 4x + 2y + 8 = 0 as (x - 2)^2 = -2(y + 2). The vertex is (2, -2) and 4a = 2, so a = 1/2. Since it opens downwards, the focus is at (2, -2 - 1/2) = (2, -5/2).

Given x = t^2 + 1 and y = 2t + 1, we have t = (y - 1)/2. Substituting into x gives x = ((y - 1)/2)^2 + 1, or (y - 1)^2 = 4(x - 1). This is a parabola with vertex (1, 1), opening right, with 4a = 4, so a = 1. The directrix is x = h - a, which is x = 1 - 1 = 0.

For the parabola y^2 = 4x, the locus of the circumcenter of triangle TPQ, where P and Q are points on the parabola and the normals at P and Q meet on the parabola, is a known property. The resulting locus is another parabola with vertex (1, 0).

A conic is the locus of a point '$P$' which moves in such a way that its distances from a fixed point'$ S $' always bears a constant ratio to its distances from a fixed straight line. 
The fixed point '$S$' is called focus. 

Given parabola may be written as,$\displaystyle \sqrt {(x-1)^2+(y-2)^2}=\frac {|x+y+3|}{\sqrt 2}$
$\Rightarrow$  focus is $S=(1,2)$, diretrix is, $x + y + 3 = 0 ....(1)$
We know axis of the parabola passes through focus and perpendicular to the directrix.
Thus equation of axis is,$x -y + 1 = 0 .....(2)$
solving (1) and (2) we get the foot of directrix $P(-2,-1)$
So the vertex of the parabola will be mid point of PS
$\Rightarrow V = \displaystyle \left (-\frac {1}{2}, \frac {1}{2}\right )$

Vertex of $y^{2} - 4y = 4x - 4a$ is $(a - 1, 2)$
So, $(a - 1 + 2 - 3)(2a - 2 + 4 - 1) < 0$
$(a - 2)(2a + 1) < 0$
$-\dfrac {1}{2} < a < 2$

${ y }^{ 2 }=4ax\quad ............(1)$

Parabola is a set of all points in a plane that are equidistant from a fixed line (called directrix) and form a fixed point called focus.

The given statement is wrong, because standard form the equation of a parabola is $y = ax^2+bx+c$, there is no cube root in equation of a parabola.

Let P(h,k) be a point on the parabola which divides the line segment joining the points A(4,-6) and B(3,1) in the ratio $\lambda:1$.
So, coordinates of point P is $\displaystyle(\frac{3\lambda+4}{\lambda+1},\frac{\lambda-6}{\lambda+1}) $
But this point lies on the parabola
$\displaystyle(\frac{\lambda-6}{\lambda+1})^{2}=4(\frac{3\lambda+4}{\lambda+1})$
$\Rightarrow (\lambda-6)^{2}=4(3\lambda+4)(\lambda+1)$
$\Rightarrow 11\lambda^{2}+40\lambda-20=0$
$\Rightarrow \displaystyle \lambda=\frac{-20\pm 2\sqrt{155}}{11}$
So, the ratio will be ${-20\pm 2\sqrt{155}}:11$

slope of axis $= \dfrac{3-1}{2+1} = \dfrac{2}{3}$
Slope of directrix $= \dfrac{-3}{2}$
Vertex is midpoint of foot of directrix and focus thus we get the coordinates oif foot of directrix as $(-4,-1)$
Equation of directrix will be $\dfrac{-3}{2}$  $= \dfrac{y+1}{x+4}$

Option A

$x^2 + 4y -6x +17 =0$

The equation of the pair of tangents is given. The directrix of a parabola is the polar of the focus with respect to the parabola. Using the properties of the pair of tangents and the focus, the directrix is calculated as 9x - 63y - 2 = 0.