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

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

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