# Any chord of the conic ${{x}^{2}}+{{y}^{2}}+xy=1$ passing through the origin is bisected at a point$(p,q)$ then $(p+q+12)$ equals to :

(a) 13

(b) 14

(c) 11

(d) 12

Answer

Verified

281.3k+ views

Hint: Make use of the property that says that the centre of a conic bisects every chord that passes through the centre of a conic section.

First, let’s find the centre of this conic section. To find this, let’s first see if the equation changes when we replace $x$ with $y$ and vice versa. If the equation remains unchanged, then that means that the centre of the conic is the origin itself.

Let’s exchange $x$ and $y$ in the equation now. Doing so, we get :

${{y}^{2}}+{{x}^{2}}+yx=1$ which is the same as the original question.

Hence, we know that the centre of this conic is in fact the origin.

It is a property of all conics that every chord drawn in the conic that passes through its centre is bisected by the centre of that conic. Applying this property to this question, since we don’t have one particular chord, and it’s any chord passing through the origin, we can say that $(p,q)$ is actually the centre of this conic. Hence, this point should be equal to the origin since that is the centre of this conic.

$\begin{align}

& (p,q)=(0,0) \\

& \Rightarrow p=0,q=0 \\

& \Rightarrow p+q+12=0+0+12=12 \\

\end{align}$

Hence, $(p+q+12)=12$ in this question.

Note:

Here, A is the centre of the conic, the origin, and as you can see, it bisects every chord that passes through it.

First, let’s find the centre of this conic section. To find this, let’s first see if the equation changes when we replace $x$ with $y$ and vice versa. If the equation remains unchanged, then that means that the centre of the conic is the origin itself.

Let’s exchange $x$ and $y$ in the equation now. Doing so, we get :

${{y}^{2}}+{{x}^{2}}+yx=1$ which is the same as the original question.

Hence, we know that the centre of this conic is in fact the origin.

It is a property of all conics that every chord drawn in the conic that passes through its centre is bisected by the centre of that conic. Applying this property to this question, since we don’t have one particular chord, and it’s any chord passing through the origin, we can say that $(p,q)$ is actually the centre of this conic. Hence, this point should be equal to the origin since that is the centre of this conic.

$\begin{align}

& (p,q)=(0,0) \\

& \Rightarrow p=0,q=0 \\

& \Rightarrow p+q+12=0+0+12=12 \\

\end{align}$

Hence, $(p+q+12)=12$ in this question.

Note:

Here, A is the centre of the conic, the origin, and as you can see, it bisects every chord that passes through it.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Scroll valve is present in a Respiratory system of class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

is known as the Land of the Rising Sun A Japan B Norway class 8 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE