JEE Advanced · Mathematics · 13. Parabola

Axis of a parabola is $y=x$ and vertex and focus are at a distance $\sqrt{2}$ and $2 \sqrt{2}$ respectively, from the origin. Then, equation of the parabola is

A
$(x-y)^2=8(x+y-2)$
B
$(x+y)^2=2(x+y-2)$
C
$(x-y)^2=4(x+y-2)$
D
$(x+y)^2=2(x-y+2)$

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Answer & Solution

Correct Answer

(A)
$(x-y)^2=8(x+y-2)$

Step-by-step Solution

Detailed explanation

As distance of vertex from origin is $\sqrt{2}$ and focus is $2 \sqrt{2}$. $\therefore V(1,1)$ and $F(2,2)$
[i.e. lying on $y=x$ ] where, length of latusrectum
\[
\begin{aligned}
& =4 a, \\
& =4 \sqrt{2}
\end{aligned}
\]
\[
[\because \text { where } a=\sqrt{2}]
\]
$\therefore$ By definition of parabola
\[
P M^2=(4 a)(P N)
\]
where, $P N$ is length of perpendicular upon $x+y-2=0$
[i.e. tangent at vertex]
\[
\begin{array}{ll}
\Rightarrow & \frac{(x-y)^2}{2}=4 \sqrt{2}\left(\frac{x+y-2}{\sqrt{2}}\right) \\
\therefore & (x-y)^2=8(x+y-2)
\end{array}
\]
Hence, the equation of parabola is $(x-y)^2=8(x+y-2)$.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A charged particle is introduced at the origin

x = 0, y = 0, z = 0

with a given initial velocity

\vec{v}

. A uniform electric field

\vec{E}

and a uniform magnetic field

\vec{B}

exist everywhere. The velocity

\vec{v}

, electric field

\vec{E}

and a uniform magnetic field

\vec{B}

are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v}$ =0	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = - B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = - B_{0} \hat{z}$

In which case will the particle describe a helical path with axis along positive z direction?

JEE Advanced 2017 Medium

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Axis of a parabola is \(y=x\) and vertex and focus are at a distance \(\sqrt{2}\) and \(2 \sqrt{2}\) respectively, from the origin. Then, equation of the parabola is

Answer & Solution

Step-by-step Solution

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