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A tangent to the ellipse (x^2)/(25)+(y^2...

A tangent to the ellipse `(x^2)/(25)+(y^2)/(16)=1` at any point `P` meets the line `x=0` at a point `Qdot` Let `R` passes through a fixed point. The fixed point is (3, 0) (b) (5, 0) (c) (0, 0) (d) (4, 0)

A

(3,0)

B

(5,0)

C

(0,0)

D

(4,0)

Text Solution

Verified by Experts

The equation of the tangent to the ellips e at `P(cos theta, 4 sin theta)` is `(xcos theta)/(5)+(y sin theta)/(4)=1`
It meets the line x=0 at Q(0,4 cosec `theta`)
The imge of Q on the line y=x is R (4 cosec `theta`, 0)
Therefore, the equation of the circle is `x(x-4"cosec"theta)+y(y-"cosec"theta)=0`
i.e., `x^(2)+y^(2)-4(x+y)"cosec" theta=0`
Therfore, each member of the family passes through intersection of `x^(2)+y^(2) =0 and x+y=0`, i.e., the point (0,0)
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