CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

An ellipse has eccentricity $$\dfrac12$$ and a focus at the point $$P \left(\dfrac12, 1\right)$$. One of its directrices is the common tangent nearer to the point $$P$$, to the circle $$\displaystyle x^2 + y^2 = 1$$ and the hyperbola $$\displaystyle x^2 - y^2 = 1$$. The equation of the ellipse is:


A
3x2+4y26x8y+4=0
loader
B
3x2+4y22x8y+4=0
loader
C
4x2+3y28x6y+4=0
loader
D
4x2+3y28x2y+4=0
loader

Solution

The correct option is B $$\displaystyle 3x^2 + 4y^2 - 2x - 8y + 4 = 0$$
There are two common tangents to the circle $$ { x }^{ 2 }+{ y }^{ 2 }=1 $$ and the hyperbola $$ { x }^{ 2 }-{ y }^{ 2 }=1. $$ These are $$ x=\pm1 $$
But $$ x=1 $$ is nearer to the point $$ \displaystyle \left( \frac { 1 }{ 2 } ,1 \right) $$ 
Therefore, directrix  of the required ellipse is $$ x=1 $$ 
Now, if $$ Q(x,y) $$ is any point on the ellipse, then its distance from the focus is 
$$ \displaystyle QP=\sqrt { { \left( x-\frac { 1 }{ 2 }  \right)  }^{ 2 }+{ \left( y-1 \right)  }^{ 2 } } $$
And its distance form the directrix is $$ \left| x-1 \right|  $$
By definition of ellipse, $$ QP=e\left| x-1 \right| $$ 
$$ \displaystyle \Rightarrow \sqrt { { \left( x-\frac { 1 }{ 2 }  \right)  }^{ 2 }+{ \left( y-2 \right)  }^{ 2 } } =\frac { 1 }{ 2 } \left| x-1 \right| $$ 
$$ \displaystyle \Rightarrow { \left( x-\frac { 1 }{ 2 }  \right)  }^{ 2 }+{ \left( y-2 \right)  }^{ 2 }=\frac { 1 }{ 4 } { \left( x-2 \right)  }^{ 2 } $$
$$ \displaystyle \Rightarrow { x }^{ 2 }-x+\frac { 1 }{ 4 } +{ y }^{ 2 }-2y+1=\frac { 1 }{ 4 } \left( { x }^{ 2 }-2x+1 \right) $$
$$ \Rightarrow { 4x }^{ 2 }-4x+1+{ 4y }^{ 2 }-8y+4={ x }^{ 2 }-2x+1 $$
$$ \Rightarrow { 3x }^{ 2 }-2x+{ 4y }^{ 2 }-8y+4=0 $$  

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image