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Condider the lines L(1):3x+4y=k-12,L(2):...

Condider the lines `L_(1):3x+4y=k-12,L_(2):3x+4y=sqrt2k` and the ellipse C :`(x^(2))/(16)+(y^(2))/(9)=1` where k is any real number
Statement-1: If line `L_(1)` is a diameter of ellipse C, then line `L_(2)` is not a tangent to the ellipse C.
Statement-2: If `L_(2)` is a diameter of ellipse C, `L_(1)` is the chord joining the negative end points of the major and minor axes of C.

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True, Statement -2 is not a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True

Text Solution

Verified by Experts

The correct Answer is:
D
We know that diameter of an ellipse passes through the centre of the ellipse. So, line `L_(1)` is a diameter of ellipse C, `k = 12` this value of K equation of line `L_(2)` becomes `3x+4y=12sqrt2 " or ", y = (-3//4)x+3sqrt2`. Clearly, it satisfies the condition of tangency i.e. `c^(2)=a^(2)m^(2)+b^(2)` of the lie `y=mx+c` and ellipse
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`.
So, statement-1 is not true.
The coordinates of the negative end-point of the major and minor axes of ellipse C are `(-4, 0) and (0, -3)` respectively. These points satisfy the equation of line `L_(1)` for k = 12. So, statement-2 is True.
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