Question

The axis of parabola is along the line y = x and the distance of vertex from origin is $\sqrt{2}$ and that of origin from its focus is $2\sqrt{2}$ . If vertex and focus both lie in the 1st 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) ${{\left( x-y \right)}^{2}}=8\left( x+y-2 \right)$

Answer 1

Hint: Parabola is symmetric about the axis of the parabola, and directrix and axis are perpendicular to each other. Distance between focus and vertex is the same as distance between vertex and directrix. Use these properties of the parabola to solve the problem. General equation of conic can be given by taking a point (x, y) on a curve, a focus and directrix of it.

Distance between (x, y) and focus = Perpendicular length of (x, y) to the directrix.

 

Complete step-by-step answer:

As the axis of the parabola is given as y = x it means focus and vertex will lie on this line, and further it is given that the distance of focus from origin is $2\sqrt{2}$ and distance of vertex from origin is $\sqrt{2}$ . And it is also given that the focus and vertex are lying in the first quadrant only. So, we can draw curves on the coordinate axis as:

As we know the axis of any parabola is perpendicular to its directrix and the distance of the directrix from vertex is the same as distance of focus of the parabola from the vertex, but directrix and focus lies opposite to the vertex of the parabola.

Now, we know the slope of the axis of the parabola is 1 (by comparing the equation with the y = mx + c form of line). Hence, y = x will form ${{45}^{\circ }}$ angle with the positive direction of axis, because slope a line is defined as the tangent of the angle formed by the line with the positive direction of x – axis i.e. $slope=\tan \theta $ , where $\theta $ is the angle of the line with the positive direction of x-axis. So, we can equate $\tan \theta $ to 1 so we get

$\tan \theta =1$

We know,

$\begin{align}

  & \tan 45=1 \\

 & \theta ={{45}^{{}}}.......................\left( i \right) \\

\end{align}$

Now, suppose A point is the vertex and B is the focus of the parabola. So, we have

$\begin{align}

  & OA=\sqrt{2} \\

 & OB=2\sqrt{2} \\

\end{align}$

Now, taking $\sin 45,\cos 45\Rightarrow \Delta OAC$ , we get

$\sin 45=\dfrac{\text{Perpendicular}}{\text{Hypotaneous}}=\dfrac{AC}{\sqrt{2}}$

We know $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ so, we get

$\begin{align}

  & \dfrac{1}{\sqrt{2}}=\dfrac{AC}{\sqrt{2}} \\

 & AC=1 \\

\end{align}$

And taking $\cos {{45}^{\circ }}$ , we get

$\cos 45=\dfrac{\text{Base}}{\text{Hypotaneous}}=\dfrac{OC}{OA}=\dfrac{OC}{\sqrt{2}}$

We know $\cos 45=\dfrac{1}{\sqrt{2}}$ , we get

$\begin{align}

  & \dfrac{1}{\sqrt{2}}=\dfrac{OC}{\sqrt{2}} \\

 & OC=1 \\

\end{align}$

Similarly, take $\sin {{45}^{\circ }},\cos {{45}^{\circ }}\Rightarrow \Delta OBD$ we get

\[\begin{align}

  & \sin {{45}^{\circ }}=\dfrac{BD}{2\sqrt{2}} \\

 & \dfrac{1}{\sqrt{2}}=\dfrac{BD}{2\sqrt{2}} \\

 & BD=2 \\

 & \Rightarrow \cos {{45}^{\circ }}=\dfrac{OD}{OB}=\dfrac{OD}{2\sqrt{2}} \\

 & \dfrac{1}{\sqrt{2}}=\dfrac{OD}{2\sqrt{2}} \\

 & OD=2 \\

\end{align}\]

So, we get coordinates of A and B as A (1, 1) and B (2, 2)

Hence, the focus of parabola is (2, 2).

As, distance between vertex and focus is

$2\sqrt{2}-\sqrt{2}=\sqrt{2}$

So, the distance between directrix and vertex will be $\sqrt{2}$ . Hence, the directrix will pass through origin and will be perpendicular to the axis of the parabola. Hence, the slope of the directrix will be ‘-1’ as we know if two lines will be perpendicular having slope as ${{m}_{1}},{{m}_{2}}$ . Then, we have

${{m}_{1}}{{m}_{2}}=-1$

So, put the slope of the axis to ‘1’ and hence the slope of the directrix is ‘-1’. Now, we know the equation of any line with slope as ‘m’ and passing through $\left( {{x}_{1}},{{y}_{1}} \right)$ can be given as

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)...................\left( ii \right)$

y – 0 = -1 (x – 0 )

y = - x or x + y = 0…………………..(iii)

Now, we know equation of any conic can be given by taking any point (x, y) on

$\sqrt{{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}}=\left| \dfrac{Ax+By+c}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|...................\left( v \right)$

Where $\left( {{x}_{1}},{{y}_{1}} \right)$ is a focus of the conic and (x, y) is a point lying on the conic and

$\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+c}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$

Is the perpendicular distance from the directrix of the point (x, y) lying on conic. The conic and use the following identity:

Distance between focus and (x, y) = Perpendicular distance of directrix from (x, y).

So, distance points $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$ can be given by distance formula as

$AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

And perpendicular distance of any point $\left( {{x}_{1}},{{y}_{1}} \right)$ from the line Ax + By + c = 0 can be given as

$\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+c}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$

So, we can get the equation of conic with the help of the above relation. So, let a point (x, y ) on the parabola and we know the focus of the parabola is (2, 2) and the equation of the directrix is x + y = 0. Hence, we get equation as

$\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}}=\left| \dfrac{x+y}{\sqrt{2}} \right|$

Squaring on both sides, we get

${{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( \dfrac{x+y}{\sqrt{2}} \right)}^{2}}$

Using the algebraic identities

$\begin{align}

  & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\

 & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\

\end{align}$

So, we get

$\begin{align}

  & {{x}^{2}}+4-2x+{{y}^{2}}+4-4y=\dfrac{{{x}^{2}}+{{y}^{2}}+2xy}{2} \\

 & 2{{x}^{2}}+2{{y}^{2}}+8-8x+x-8y={{x}^{2}}+{{y}^{2}}+2xy \\

 & {{x}^{2}}+{{y}^{2}}-2xy=8\left( x+y-2 \right) \\

\end{align}$

Hence, option (d) is correct.

 

Note: one may go wrong if he or she suppose the parabola of the form of

$\begin{align}

  & {{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right) \\

 & \Rightarrow {{\left( x-{{x}_{1}} \right)}^{2}}=4a\left( y-{{y}_{1}} \right) \\

\end{align}$

As the axis of these parabola are x and y axis respectively. So, don’t use these equations to solve the problem.

We don’t need to use the trigonometric function for the $\angle {{45}^{\circ }}$ . As one may suppose vertex and focus as (a, a) and (b, b) and hence applying the Pythagoras in $\Delta OAC,\Delta OBD$ . It means we get equations as

${{a}^{2}}+{{a}^{2}}={{\left( \sqrt{2} \right)}^{2}}=2,{{b}^{2}}+{{b}^{2}}={{\left( 2\sqrt{2} \right)}^{2}}=8$

Be clear with the property that the axis of the parabola and directrix are perpendicular to each other and remember the relation of the general equation of conic sections, with the help of which we need to calculate the equation of the curves.