Question

Let $$E$$ be the ellipse $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$$   and $$C$$ be the circle $${x^2} + {y^2} = 9.$$   Let $$P = \left( {1,\,2} \right)$$   and $$Q = \left( {2,\,1} \right).$$   Which one of the following is correct ?

A. $$Q$$ lies inside $$C$$ but outside $$E$$
B. $$Q$$ lies outside both $$C$$ and $$E$$
C. $$P$$ lies inside both $$C$$ and $$E$$
D. $$P$$ lies inside $$C$$ but outside $$E$$  
Answer :   $$P$$ lies inside $$C$$ but outside $$E$$
Solution :
Given equation of ellipse $$E$$ is
$$\eqalign{ & \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1 \cr & \Rightarrow \frac{{4{x^2} + 9{y^2}}}{{36}} = 1 \cr & \Rightarrow 4{x^2} + 9{y^2} = 36 \cr & \Rightarrow 4{x^2} + 9{y^2} - 36 = 0......\left( 1 \right) \cr} $$
And $$C\,:$$  equation of circle is $${x^2} + {y^2} = 9$$
Which can be rewritten as $${x^2} + {y^2} - 9 = 0......\left( 2 \right)$$
For a point $$P = \left( {1,\,2} \right)$$   we have
$$\eqalign{ & 4{\left( 1 \right)^2} + 9{\left( 2 \right)^2} - 36 = 40 - 36 > 0\,\,\,\,\left[ {{\text{from equation }}\left( 1 \right)} \right] \cr & \,{\text{and }}\,{1^2} + {2^2} - 9 = 5 - 9 < 0\,\,\,\,\,\left[ {{\text{from equation }}\left( 2 \right)} \right] \cr} $$
$$\therefore $$  Point $$P$$ lies outside of $$E$$ and inside of $$C.$$

Releted MCQ Question on
Geometry >> Ellipse

Releted Question 1

Let $$E$$ be the ellipse $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$$   and $$C$$ be the circle $${x^2} + {y^2} = 9.$$   Let $$P$$ and $$Q$$ be the points $$\left( {1,\,2} \right)$$  and $$\left( {2,\,1} \right)$$  respectively. Then-

A. $$Q$$ lies inside $$C$$ but outside $$E$$
B. $$Q$$ lies outside both $$C$$ and $$E$$
C. $$P$$ lies inside both $$C$$ and $$E$$
D. $$P$$ lies inside $$C$$ but outside $$E$$
Releted Question 2

The radius of the circle passing through the foci of the ellipse $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1,$$   and having its centre at $$\left( {0,\,3} \right)$$  is-

A. $$4$$
B. $$3$$
C. $$\sqrt {\frac{1}{2}} $$
D. $$\frac{7}{2}$$
Releted Question 3

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1,$$    is-

A. $$\frac{{27}}{4}\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
B. $$9\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
C. $$\frac{{27}}{2}\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
D. $$27\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
Releted Question 4

If tangents are drawn to the ellipse $${x^2} + 2{y^2} = 2,$$   then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is-

A. $$\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1$$
B. $$\frac{1}{{4{x^2}}} + \frac{1}{{2{y^2}}} = 1$$
C. $$\frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1$$
D. $$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{2} = 1$$

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Ellipse


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