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Question

# The line 3x+4y=√7 touches the ellipse 3x2+4y2=1 at the point

A
(7,7)
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B
(17,17)
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C
(17,17)
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D
(7,7)
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Solution

## The correct option is B (1√7,1√7)Let the given line touches the ellipse at point P(θ)=(cosθ√3,sinθ2) The equation of the tangent at P is √3xcosθ+2ysinθ=1…(i) Comparing equation (i) with the given equation of the line 3x+4y=√7, we get ⇒cosθ√3=sinθ2=1√7 ∴ Point of contact is (1√7,1√7) Alternate Solution: For the given ellipse a2=1/3,b2=1/4 Given tangent is 3x+4y=√7⇒y=−34x+√74 ∴m=−34,c=√74 Using slope form the point of contact will be (−a2mc,b2c)≡(1√7,1√7)

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