Answer

Verified

403.2k+ views

**Hint:**Equation of chord of contact of tangent from point $({{x}_{1}},{{y}_{1}})$ where $({{x}_{1}},{{y}_{1}})$ lies outside the ellipse is $\dfrac{x{{x}_{1}}^{{}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$ . This equation of chord is the tangent to the circle. So if $y=mx+c$ where m is the slope of line and c is the y-intercept, is the tangent to the circle ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ with radius $a$ then the radius of circle is given by $a=\dfrac{\left| c \right|}{\sqrt{1+{{m}^{2}}}}$.

**Complete step by step answer:**

Standard equation of circle is ${{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$ where $(a,b)$ is the center of the circle and $r$ is the radius of the given circle.

In the given question we are given a circle ${{x}^{2}}+{{y}^{2}}=100$ , whose center is (0,0) and radius = 10. And $\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=1$ is a given ellipse having major axis in the y-axis direction. Graph of the circle and ellipse is given below.

Equation of chord of contact of tangent from point $({{x}_{1}},{{y}_{1}})$ where $({{x}_{1}},{{y}_{1}})$ lies outside the ellipse$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\dfrac{x{{x}_{1}}^{{}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1$.Let $(h,k)$ be the point outside the ellipse then equation of chord of contact of tangent from point A$(h,k)$ to ellipse $\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{9}=1$ is $\dfrac{hx}{4}+\dfrac{ky}{9}=1.............(1)$.

This chord is the tangent to the circle. If $y=mx+c...........(2)$ where m is the slope of line and c is the y-intercept, is the tangent to the circle ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ with radius $a$ then the radius of circle is given by $a=\dfrac{\left| c \right|}{\sqrt{1+{{m}^{2}}}}............(3)$.

Let $p=\dfrac{h}{4}$ ,$q=\dfrac{k}{9}$ and $r=1$ then equation (1) can be written as

\[\Rightarrow px+qy=r\].

$\Rightarrow qy=r-px$.

$\Rightarrow y=\dfrac{r}{q}-\dfrac{px}{q}$.

Comparing equation (2) and (4) we get $m=\dfrac{-p}{q}$ and $c=\dfrac{r}{q}$. Putting these value in equation (3) we get $\begin{align}

& a=\dfrac{\left| \left( \dfrac{r}{q} \right) \right|}{\sqrt{1+{{\left( \dfrac{-p}{q} \right)}^{2}}}} \\

& \\

\end{align}$.

Squaring on both sides we get,

$\Rightarrow {{a}^{2}}={{\left( \dfrac{\left| \left( \dfrac{r}{q} \right) \right|}{\sqrt{1+{{\left( \dfrac{-p}{q} \right)}^{2}}}} \right)}^{2}}$.

\[\Rightarrow {{a}^{2}}=\dfrac{{{\left( \dfrac{r}{q} \right)}^{2}}}{1+{{\left( \dfrac{-p}{q} \right)}^{2}}}\].

$\Rightarrow {{a}^{2}}=\dfrac{1}{\left( \dfrac{1+{{\left( \dfrac{-p}{q} \right)}^{2}}}{{{\left( \dfrac{r}{q} \right)}^{2}}} \right)}$.

$\Rightarrow {{a}^{2}}=\dfrac{1}{{{\left( \dfrac{q}{r} \right)}^{2}}+{{\left( \dfrac{p}{r} \right)}^{2}}}...............(5)$.

Here $r=1$ so equation (5) becomes ${{a}^{2}}=\dfrac{1}{{{q}^{2}}+{{p}^{2}}}...........(6)$.

We know that the radius of circle ${{x}^{2}}+{{y}^{2}}=100$ is 10 so here $a$ is 10. Putting the values of $a=10$, $p=\dfrac{h}{4}$ and$q=\dfrac{k}{9}$ in equation (6) we get

$\Rightarrow 100=\left( \dfrac{1}{\dfrac{{{h}^{2}}}{16}+\dfrac{{{k}^{2}}}{81}} \right)$.

$\Rightarrow \dfrac{{{h}^{2}}}{16}+\dfrac{{{k}^{2}}}{81}=\dfrac{1}{100}$.

$\Rightarrow \dfrac{100{{h}^{2}}}{16}+\dfrac{100{{k}^{2}}}{81}=1$.

$\Rightarrow \dfrac{{{h}^{2}}}{\left( \dfrac{16}{100} \right)}+\dfrac{{{k}^{2}}}{\left( \dfrac{81}{100} \right)}=1.............(7)$.

Standard equation of ellipse is given as $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ here we got $b < a$, comparing this equation with (7) we get ${{b}^{2}}=\dfrac{81}{100}$ and ${{a}^{2}}=\dfrac{16}{100}$.

Eccentricity is given by formula, $e=\dfrac{c}{b}..........(8)$, where $c=\sqrt{{{b}^{2}}-{{a}^{2}}}$.

$\Rightarrow c=\sqrt{\dfrac{81}{100}-\dfrac{16}{100}}$.

$\Rightarrow c=\sqrt{\dfrac{81-16}{100}}$.

$\Rightarrow c=\dfrac{\sqrt{65}}{10}$.

Putting the value of c and a in equation (8) we get,

$\Rightarrow e=\dfrac{\left( \dfrac{\sqrt{65}}{10} \right)}{\left( \dfrac{9}{10} \right)}$.

$\Rightarrow e=\dfrac{\sqrt{65}}{9}$.

Therefore,

$\dfrac{81{{e}^{2}}}{13}=\dfrac{81}{13}\times \dfrac{65}{81}$.

$\dfrac{81{{e}^{2}}}{13}=\dfrac{65}{13}$.

$\dfrac{81{{e}^{2}}}{13}=5$.

**Hence the value of $\dfrac{81{{e}^{2}}}{13}$ is 5.**

**Note:**

We need not always get the equation of ellipse as $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ where $a>b$. We should know that the equation of the ellipse depends on the condition given in the problem. We should not confuse it with eccentricity while solving this problem. We can also solve this problem by taking the equation of common chords at the points $\left( 0,10 \right)$ and $\left( 10,0 \right)$ of the points which gives us the values of a and b directly.

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Name 10 Living and Non living things class 9 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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

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

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail