Question

# A tangent $$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$$ meets the axes at $$A$$ and $$B$$. Then the locus of mid point of $$AB$$ is

A
a2x2+b2y2=2
B
a2x2+b2y2=4
C
a2x2+b2y2=1
D
a2x2+b2y2=12

Solution

## The correct option is B $$\cfrac { { a }^{ 2 } }{ { x }^{ 2 } } +\cfrac { { b }^{ 2 } }{ { y }^{ 2 } } =4$$Tangent to the ellipse is given by,$$\dfrac{xcosm\theta}{a}+\dfrac{ysin\theta}{b}=1$$Since this line intersect the cordinate axisat $$y=0,\dfrac{xcos\theta}{a}=1\implies x=\dfrac{a}{cos\theta}$$so the cordinate is $$(\dfrac{a}{cos\theta},0)$$Similarlyat $$x=0,\dfrac{ysin\theta}{b}=1\implies y=\dfrac{b}{sin\theta}$$Cordinate is $$(0,\dfrac{b}{sin\theta})$$Midpoint is given by-$$(h,k)=\dfrac{\dfrac{a}{cos\theta}+0}{2},\dfrac{0+\dfrac{b}{sin\theta}}{2}$$$$(h,k)=(\dfrac{a}{2cos\theta},\dfrac{b}{2sin\theta})$$we get.$$cos\theta=\dfrac{a}{3h},sin\theta=\dfrac{b}{2k}$$As we know$$sin^2\theta+cos^2\theta=1\\\implies \dfrac{a^2}{4h^2}+\dfrac{b^2}{4k^2}=1$$$$\dfrac{a^2}{h^2}+\dfrac{b^2}{k^2}=4$$therefore the locus of midpoint  is$$\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}=4$$Mathematics

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