**Theorem:** Three normal can be drawn from a point (x₁, y₁) to the parabola y² = 4ax.

If the normal at t₁ and t₂ to the parabola y² = 4ax meet on the parabola, then t₁t₂ = 2.

**Proof: **Let the normal at t₁ and t₂ meet at t₃ on the parabola

The equation of the normal at t₁ is y + xt₁ = 2at₁ + at₁³ … (1)

Equation of the chord joining t₁ and t₃ is y (t₁ + t₃) = 2x + 2at₁t₃ … (2)

from equation (1) and (2) represent the same line

Therefore \(\frac{{{t}_{1}}+{{t}_{3}}}{1}=\frac{-2}{{{t}_{1}}}\),

\({{t}_{3}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\),

Similarly, t₃ = -t₂ – 2/t₂

Therefore \(-{{t}_{1}}-\frac{2}{{{t}_{1}}}=-{{t}_{2}}-\frac{2}{{{t}_{2}}}\),

\({{t}_{1}}-{{t}_{2}}=\frac{2}{{{t}_{2}}}-\frac{2}{{{t}_{1}}}\),

\({{t}_{1}}-{{t}_{2}}=\frac{2({{t}_{1}}-{{t}_{2}})}{{{t}_{1}}{{t}_{2}}}\),

t₁ t₂ = 2

**Example: **Show that the tangent at one extremity of a focal chord of a parabola is parallel to the normal at the other extremity.

**Solution: **P(t₁), Q(t₂) are the ends of a focal chord

Slope of PS = Slope of PQ

\(\frac{2a{{t}_{1}}}{a({{t}_{1}}^{2}-1)}=\frac{2a({{t}_{1}}-{{t}_{2}})}{a({{t}_{1}}^{2}-{{t}_{2}}^{2})}\),

\(\frac{{{t}_{1}}}{{{t}_{1}}^{2}-1}=\frac{1}{{{t}_{1}}+{{t}_{2}}}\),

\({{t}_{1}}+{{t}_{2}}=\frac{{{t}_{1}}^{2}-1}{{{t}_{1}}}={{t}_{1}}-\frac{1}{{{t}_{1}}}\),

\({{t}_{2}}=\frac{-1}{{{t}_{1}}}\)… (1)

Equation of the tangent at P(t₁) is t₁y = x + at₂²

slope of the tangent at P = 1/t₁ … (2)

y + xt₂ = 2at₂ + at₂²

Slope of the normal at Q = -t₂ … (3)

From equation (1), (2), (3) we get

Slope of the tangent at P = slope of normal at Q.

Slope of the tangent at P is parallel to the normal at Q.

]]>**What is Velocity?**

Velocity is the Rate of change of position. I.e. Rate of displacement with time is called velocity. A velocity of an object is the rate of change of the object’s position with respect to a frame of reference and time, it might sound complicated but velocity is basically speed in a specific direction. It is a vector quantity, which means we need both magnitude and direction to define velocity. The SI unit of it is meter per second (m/ sec) if there is a change in magnitude or the direction in velocity of a body the body is said to be accelerating.

**Types of Velocity:**

**1. Uniform Velocity: **A particle is said to have uniform velocity, if magnitudes as well as direction of its velocity remains same and this is possible only when the particles moves in same straight line without reversing its direction.

**2. Non – Uniform Velocity: **A particle is said to have non – uniform velocity, if either of magnitude or direction of velocity changes.

**3. Average Velocity: **It is defined as the ratio of displacement to time taken by the body.

\(Average\,\,Velocity({{\overrightarrow{v}}_{avg}})\,\,=\,\,\frac{Displacement}{Time\,\,Taken}\).

\({{\overrightarrow{v}}_{avg}}\,\,=\,\,\frac{\Delta \overrightarrow{r}}{\Delta t}\).

**4. Instantaneous Velocity: **Instantaneous velocity is defined as rate of change of position vector of particles with time at a certain instant of time.

Instantaneous Velocity \(\overrightarrow{v}\,\,=\,\,\underset{t\to 0}{\mathop{\lim }}\,\,\frac{\Lambda \overrightarrow{r}}{\Delta t}\,\,=\,\,\frac{d\overrightarrow{r}}{dt}\).

**How to find Velocity?**

**Problem: **A car travels at uniform velocity a distance of 100 m in 4 seconds. What is the velocity of the car?

**Solution: **Given,

Distance (d) = 100 m

Time (t) = 4 sec

Velocity of the Car (V) =?

We know that:

\(Velocity(V)\,\,=\,\,\frac{Dis\tan ce(d)}{time(t)}\).

\(Velocity(V)\,\,=\,\,\frac{100\,m}{4\,\sec }\,\,=\,\,25\,m/\sec \).

]]>A Photodiode is a p – n junction diode that can absorb photons and generate either a Photo Voltage or Free Carriers that can produce Photocurrent. They are used for detection of Optical signals and for conversion of Optical Power to Electrical Power.

A Photodiode is an example of an optoelectronic junction device, which implies that it is used as an electrical to optical or an optical to electrical transducer. It works on the effect of light falling onto a diode which leads to the generation of current through it. It is implemented over a special p-n junction diode by fabricating a transparent window on top of it which allows light to be incident on the diode.

**Photodiode Symbol: **The symbol of photodiode is similar to the normal p-n junction diode except that it contains arrows striking the diode. The arrows striking the diode represent light or photons.

A photodiode has two terminals: a cathode and an anode.

A photodiode is subjected to photos in the form of light which affects the generation of electron–hole pairs. If the energy of the falling photons (hv) is greater than the energy gap (E_{g}) of the semiconductor material, electron-hole pairs are created near the depletion region of the diode. The electron-hole pairs created are separated from each other before recombining due to the electric field of the junction. The direction of electric field in the diode forces the electrons move towards the n – side and consequently the holes move towards the p – side, a rise in the electromotive force is observed. Now when an external load is connected to the system, a current flow is observed through it.

The more the electromotive force created, the greater is the current flow. The magnitude of the electromotive force created depends directly upon the intensity of the incident light. This effect of proportional change in photocurrent with change in light intensity can be easily observed by applying a reverse bias.

Since photodiodes generate current flow directly depending upon the light intensity received, they can be used as photo detectors to detect optical signals. Built-in lenses and optical filters may be used to enhance the power and productivity of a photodiode.

**Types of photodiodes: **The working operation of all types of photodiodes is same. Different types of photodiodes are developed based on specific application. For example, PIN photodiodes are developed to increase the response speed. PIN photodiodes are used where high response speed is needed.

The different types of photodiodes are:

- PN junction photodiode
- PIN photodiode
- Avalanche photodiode

Among all the three photodiodes, PN junction and PIN photodiodes are most widely used.

]]>DATE | 10.30 am to 1.30 pm Reporting 10.00 am Rank Ranges (Based on All India Rank) |
2.30 pm to 4.30 pm Reporting 2.00 pm Rank Ranges (Based on All India Rank) |

19-07-2018 |
1 to 2,00,000 | 2,00,001 to 4,00,000 |

20-07-2018 | 400001 to 634897 (Non-Karnataka) 400001 to 757569 (Karnataka) |
– |

Length of a vector in term of in components

If r = xi + yj + zk then |r| = \(\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\).

**Proof:** Let \(r=\overrightarrow{OP}\) then P = (x, y, z)

Now |r| = OP = \(\sqrt{{{(x-0)}^{2}}+{{(y-0)}^{2}}+{{(z-0)}^{2}}}\),

OP = \(\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\).

**Examples: **r = 2i + 3j – 6k then find |r|

**Solution: **Given that,

r = 2i + 3j – 6k

Now |r| = \(\sqrt{{{(2-0)}^{2}}+{{(3-0)}^{2}}+{{(6-0)}^{2}}}\),

= \(\sqrt{{{(2)}^{2}}+{{(3)}^{2}}+{{(6)}^{2}}}\),

= \(\sqrt{4+9+36}\),

= \(\sqrt{49}\),

= 7

**Division formulas **

- Let a, b be the position vectors of the point A, B respectively. The position vectors of the point P which divided \(\overline{AB}\) in the ratio m : n is \(\frac{mb+na}{m+n}\). Conversely the point P with the position vector \(\frac{mb+na}{m+n}\) lies on the \(\overline{AB}\) and divides \(\overline{AB}\) in the ratio m : n.
- Let a, b be the position vectors of the point A, B respectively. The position vector of the point P which divides \(\overline{AB}\) externally in the ratio m : n in \(\frac{mb-na}{m-n}\).
- The point which divides the line segment joining (x₁, y₁, z₁) and (x₂, y₂, k₂) ration m : n is internally is \(\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\frac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)\).
- The point which divides the line segment joining (x₁ , y₁ , z₁) and (x₂, y₂, k₂) ration m : n externally is \(\left( \frac{m{{x}_{2}}-n{{x}_{1}}}{m-n},\frac{m{{y}_{2}}-n{{y}_{1}}}{m-n},\frac{m{{z}_{2}}-n{{z}_{1}}}{m-n} \right)\).
- The ratio in which xi + yj + zk divides the line segment joining x₁i + y₁j + z₁k and x₂i + y₂j + z₂k is x – x₁ : x₂ – x = y – y₁ : y₂ – y = z – z₁ : z₂ – z.

**Example: **Find the position vector of the point which divided the line joining the point 3a – 2b and a + b in the ratio 2 : 1

1. Internally and

2. Externally

**Solution: **Given that the line joining the point 3a – 2b and a + b in the ratio 2 : 1

We know that internally divided in ratio \(\frac{mb+na}{m+n}\),

Externally divided in ratio \(\frac{mb-na}{m-n}\),

Position vector the point which divides the line joining the given point in the line ratio 2 : 1 Internally is

= \(\frac{2(a+b)+1(3a-2b)}{2+1}\),

= \(\frac{5a}{3}\),

Position vector the point which divides the line joining the given point in the line ratio 2 : 1 Externally is

= \(\frac{2(a+b)-1(3a-2b)}{2-1}\),

= – a + 4b.

]]>Nuclear force is one of the four fundamental forces of nature, the others being gravitational and electromagnetic forces. In fact, being 10 million times stronger than the chemical binding forces, they are also known as the strong forces. The nuclear force acts between all of the particles in the nucleus, i.e., between two neutrons, between two protons, and between a neutron and a proton. It is attractive in all cases. In contrast, an electrical force acts only between two protons and it is repulsive. The nucleus is held by the forces which protect them from the enormous repulsion forces of the positive protons. It is a force with short range and not similar to the electromagnetic force. We know that the nucleus is made up with its fundamental particles that are the protons and neutrons.

**What is Nuclear force?**

Nuclear force is the force that binds the proton and neutrons in a nucleus together. This force can exist between protons and protons, neutrons and protons or neutrons and neutrons. This force is what holds the nucleus together. The charge of protons, which is +1e tends to push them away from each other with a strong electric field repulsive force. But nuclear force is strong enough to keep them together and to overcome that resistance at short range.

**Characteristics of Nuclear force:**

**Nuclear forces are attractive in nature:**The magnitude which depends upon inter nucleon distance is of very high order.**Nuclear forces are charge Independent:**Nature of force remains the same whether we consider force between two protons, between two neutrons or between a proton and a neutron.**These are Short Range forces:**Nuclear forces operate between two nucleons situated in close neighbourhood only.**Nuclear forces decrease very quickly with distance between two nucleons:**Their rate of decrease is much rapid than that of inverse square law forces. The forces become negligible when the nucleons are more than 10^{-12 }cm apart.**Nuclear force are spin dependent:**Nucleons having parallel spin are more strongly bound to each other than those having anti-parallel spin.

**Examples: **The binding of protons, which are repulsive in nature because of their positive charge. On a larger scale, this force is responsible for the immense destructive power of nuclear weapons. It is also used in Nuclear power plants to generate heat for the purpose of generating energy, such as electricity.

ADMISSION NOTIFICATION FOR Bachelor of Physiotherapy (BPT), B.Sc., (Medical Laboratory Technology), B.Sc., (Nursing) 4 YDC and Post Basic B.Sc., (Nursing) 2 YDC COURSES FOR THE ACADEMIC YEAR – 2018-19 under competent Authority Quota Seats in the State of AP.

The candidate has to apply separate application form for each course

- Bachelor of Physiotherapy i.e., BPT
- Bachelor of Science in Medical Laboratory Technology i.e., B.Sc., (MLT)
- Bachelor of Science in Nursing i.e., B.Sc., (Nursing) 4Years Degree course
- Post Basic Nursing i.e., B.Sc., (Nursing) 2 Years Degree course.

**Eligibility for BPT Course:** Intermediate (10+2) or equivalent with Botany, Zoology, Physics and Chemistry or Intermediate Vocational (Physiotherapy) or Inter vocational with Bridge course of Biological and Physical Sciences or APOSS with Biological and Physical Sciences and should have completed 17 years of age as on 31.12.2018.

**Eligibility for B.Sc., (MLT) Course:** Intermediate (10+2) or equivalent with Botany, Zoology, Physics and Chemistry or Intermediate Vocational (Medical Lab Technology) or Diploma in Medical Laboratory Technology or Inter vocational with Bridge course of Biological and Physical Sciences or APOSS with Biological and Physical Sciences and should have completed 17 years of age as on 31.12.2018

**Eligibility for B.Sc., (Nursing) 4 YD Course:**

**1.** The minimum age for admission shall be 17 years on 31st December of the year in which admission is sought.

**2. Minimum Education:**

a. 10+2 Passed @45% aggregate of marks in science (Physics, Chemistry and Biology) and English is compulsory subject.

b. 10+2 passed out from recognized board under AISSCE/ CBSE/ ICSE/ SSCE/ HSCE/ NIOS/ APOSS or other equivalent Board with Physics, Chemistry and Biology.

c. Intermediate Vocational with Bridge Course in Biological and Physical Sciences.

d. Those who belongs to SC, ST and BC are eligible with 40% in the Science Group subjects and English is compulsory subject as per orders of the Secretary Indian Nursing Council, New Delhi vide letter No.F.No.1-2/2014-INC and vide the circular No. F.NO.1-6/2016-INC, Dated.21-09-2016.

**3.** Students shall be medically fit.

**4.** Students shall be admitted once in a year (As per the G.O.Rt.No.1095, Health Medical and Family Welfare (K1) Depart., dated.02.09.2013)

**Age:** Candidates born on or after 02.01.2002 are not eligible for admission into the above courses for underage verification

**Eligibility for Post B.Sc., (Nursing) 2YD Course:** The candidates should be of Indian Nationality or persons of Indian origin (P.I.O)/ Overseas citizen of India (OCI) card holders and should satisfy the local or non-local status in Andhra Pradesh (Residence requirement) as laid down in Andhra Pradesh educational institutions (Regulations of admissions) order, 1974 and the selection will be done as per the procedure laid down in the G.O.P.No.646, dated 10-07-1979 as amended in G.O.Ms.No.42, Higher Education (EC2) Dept., dated 18-05-2009.

Male and female Candidates are eligible for admission into Post Basic B.Sc., (N) 2 Years Course as per 228^{th} EC resolution dated 05.05.2018.

** Education Qualifications:** Candidates seeking admission into Post Basic B.Sc., (Nursing) Course should have passed Intermediate Examination.

or

Any other equivalent examination (10+2 pattern) recognized by the University in A.P. or Board of Intermediate Education Andhra Pradesh or NIOS or APOSS – Indian Nursing Council Regulations.

and

Should have passed General Nursing and Midwifery from an institution recognized by the Govt. of A.P. or Nursing Council of India and shall be registered with the State Nursing Council.

(Eligibility Criteria is as per G.O.Ms.No.269, HM & FW (K2) Dept., Dt. 30-11-2009 as amended in G.O.Ms.No.194, HM & FW (K2) Dept., Dt. 07-07-2011, G.O.Ms.No.40, HM & FW (K2) Dept., Dt.18-02-2012 and G.O.Ms.No.33, HM & FW (K1) Dept., Dt.05-03-2013 and from time to time.

** Government Service Candidates:** Should have completed 2 years of service as a staff nurse in Medical and Health Department and Andhra Pradesh Vaidya Vidhana Parishad as on 30.06.2018. Service certificate as a staff nurse issued by the respective employer should be enclosed. Must be a regular employee. Service certificate and regularization orders as staff nurse should be produced at the time of original certificate verification.

** Age:** Candidate should have completed 17 years of age and should not be more than 45 years of age as on 31st December of the year of admission. In case of SC/ ST Candidates the maximum age shall be relaxed by 3 years.

**Registration and Verification Fee: **For each course separately

Rs. 2,000/- for OC

Rs. 1,600/- for BC/SC/ST Candidates

**IMPORTANT DATES:**

1 | Publication of notification in Press | 13.07.2018 |

2 | Application form available website: https://paramed.apntruhs.in | 17.07.2018from 11.00 AM to 06.08 .2018 by 05.00 PM |

3 | Dates of verification of original certificates | To be notified |

4 | Display of merit list | To be notified |

5 | Dates of exercising web options | To be notified |

**FEE PARTICULARS:**

** Tuition Fee:** The Tuition fee is as per G.O.Ms.No.153, H.M & F.W (C2) Dept., dated.12-09-2017.

** University Fee: **University fee: All the selected candidates shall pay the following fee i.e., Rs.5000/- at the time of downloading of allotment order through online. University fee once paid shall not be refunded under any circumstances.

1 | Registration Fee | Rs. 2000/- |

2 | Admission Fee | Rs. 1000/- |

3 | IUT Fee/ Sports Fee | Rs. 800/- |

4 | Students Welfare Fund | Rs. 400/- |

5 | Hand Book | Rs. 200/- |

6 | Digital Library Fee | Rs. 600/- |

Total | Rs. 5000/- |

All the selected candidates have to pay the fee and other constituent fees, special fees and caution money deposits at the rates laid down for the purpose from time to time.

]]>Find the periods for the given functions.

1. Cos (3x + 5) + 7

**Solution:** Given that

Cos (3x + 5) + 7

Period of cosx = 2π

Period = \(\frac{period\ of\ \cos x}{[coefficient\ of\ x]}=\frac{2\pi }{3}\)

2. Tan5x

**Solution: **Given that

Tan5x

f(x) = tan5x

Period of tanx = π

Period = \(\frac{period\ of\ \cos x}{[coefficient\ of\ x]}=\frac{\pi }{5}\)

3. |sinx|

**Solution: **Given that,

|sinx|

Period = π

|sin(π + x)| = |(-sinx)| = sinx

4. tan (x + 4x + 9x + …… + n²x) (n any positive integer)

**Solution: **Given that

tan (x + 4x + 9x + …… + n²x)

1 + 2² + 3² + …… + n² = \(\frac{n(n+1)(2n+1)}{6}\)

tan(1 + 4 + 9 + …… + n²)x

tan[n(n+1)(2n+1)/6]x

period = \(\frac{\pi }{\frac{n(n+1)(2n+1)}{6}}=\frac{6\pi }{n(n+1)(2n+1)}\)

5. sketch the region enclosed by y = sinx, y = cosx and x – axis in the interval [0, π]

**Solution: **y = sinx

X |
0 | π/4 | π/2 | 3π/4 | π |

Y | 0 | 1/√2 | 1 | 1/√2 |
0 |

y = cosx

X |
0 | π/4 | π/2 | 3π/4 | π |

Y | 1 | 1/√2 | 0 | -1/√2 |
-1 |

A light-emitting diode (LED) is a semiconductor device that emits visible light when an electric current passes through it. The light is not particularly bright, but in most LEDs it is monochromatic, occurring at a single wavelength. The output from an LED can range from red to blue-violet. Some LEDs emit infrared (IR) energy, such a device is known as an infrared-emitting diode (IRED).

A light emitting diode (LED) is an example of an Optoelectronic junction device, which implies that it is used as an electrical to optical or an optical to electrical transducer. It works on the effect of suitable voltage being applied to the diode which leads to the generation of light in the form of photons. It is implemented over a heavily doped p-n junction diode by fabricating a transparent window on top of it which allows the generated light to be emitted out of the diode.

When a forward bias voltage is applied to a diode both electrons and holes move in opposite directions where they are minority carriers i.e. electrons move from n-side to p-side and holes move from p-side to n-side. Due to this movement, the equilibrium is disturbed and consequently the concentration of minority carriers at the boundary of the junction increases. Because of this excess presence of minority carriers at the boundary junction, they combine with majority carriers and this combination releases energy in the form of photons. The photons that are emitted have energy equal to or greater than the band gap energy for the diode and hence they are able to escape from it in the form of light.

The intensity of the light emitted is directly proportional to the intensity of the bias voltage being applied to the diode. However after reaching a peak value, on further increase in the bias voltage the emitted light intensity decreases. Therefore it is necessary to apply the bias in such a manner that maximum efficiency is obtained.

]]>When a positive charge q is placed on a conductor that is insulated from ground, an electric field emanates from the conductor to ground and the conductor will have a nonzero potential V relative to grounds. If more charges is placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance C of this conductor.

**Capacitance of a Capacitor: **Capacitance is the property of the capacitor that defines the maximum amount of electrical charge stored in it. Capacitance may vary depending on the shape of the capacitor. It can be calculated by using the geometry of the conductors and dielectric material properties.

Capacitance is defined as the ratio of charge (Q) on the either plates to the potential difference (V) between them.

Capacitance \((C)\,\,=\,\,\frac{Q}{V}\)

Current can be obtained as,

\(I(t)\,\,=\,\,C\,\left[ \frac{d(V)}{d(t)} \right]\)

So, Capacitance is directly proportional to the charge (Q) and is inversely proportional to the voltage (V).

Capacitance of the capacitor can be increased by increasing the number of plates.

**Capacitance Formula: **Capacitance is to describe how much charge any conductor can hold. It is the ratio of the charge flowing across the conductor to the potential applied. Capacitance is the ability of a substance to store an electrical charge. A parallel plate capacitor is the common form of energy storage device. Capacitance is exhibited by a parallel plate arrangement and defined in terms of charge storage.

Capacitance formula can be expressed as:

\(C\,\,=\,\,\frac{Q}{V}\)

Where,

Q = Charge of the Conductor.

V = Potential applied across the conductor.

C = Proportionality Constant or Capacitance.

**Standard units of Capacitance:**

Farads is a high value so, capacitance is expressed as sub units of capacitor real time such as micro farads (μF), nano farads (nF) and pico farads (PF).

Most of the electrical applications are covered by the following standard unit (SI) prefixes for easy calculations,

1 mF (milli Farad) = 10⁻³ F = 1000 μF = 1000000 nF

1 μF (micro Farad) = 10⁻⁶ F = 1000 nF = 1000000 PF

1 nF (nano Farad) = 10⁻⁹ F = 1000 PF

1 PF (Pico Farad) = 10⁻¹² F

To convert μF to nF or PF to a wide range of other units and vice versa, we need to use Electric Capacitance Unit Converter.

]]>